Why does the series $sum_{n=1}^inftyfrac1n$ not converge?











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Can someone give a simple explanation as to why the harmonic series




$$sum_{n=1}^inftyfrac1n=frac 1 1 + frac 12 + frac 13 + cdots $$




doesn't converge, on the other hand it grows very slowly?



I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.










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  • 3




    This is not meant to be an answer but an interesting note. Suppose we denote $H(n) = 1/1 + 1/2 + ... + 1/n$ then $H(n!) - H((n-1)!) approx log(n)$ for large n. Does this give a hint? ;)
    – Roupam Ghosh
    Jul 11 '11 at 4:14






  • 5




    Here is a weakly related question: What is a textbook, or even a popularization for the general public, that (1) discusses infinite series, but (2) does not have an explanation for the divergence of this exact series?
    – GEdgar
    Nov 3 '13 at 19:50










  • to avoid defining the logarithm, use the Cauchy condensation test to show that $sum 1/n$ converges iff $sum 1$ converges
    – reuns
    Jan 30 '16 at 23:31












  • These are two of my favourite papers: The Harmonic Series Diverges Again and Again and More Proofs of Divergence of the Harmonic Series. See these.
    – user249332
    Mar 9 '16 at 22:56










  • If it converges, then it contradicts the dominated convergence theorem. This proof is easily comprehensible if you know the dominated convergence theorem, but that theorem is not the most comprehensible.
    – Oiler
    Oct 6 '16 at 23:08

















up vote
120
down vote

favorite
54












Can someone give a simple explanation as to why the harmonic series




$$sum_{n=1}^inftyfrac1n=frac 1 1 + frac 12 + frac 13 + cdots $$




doesn't converge, on the other hand it grows very slowly?



I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.










share|cite|improve this question




















  • 3




    This is not meant to be an answer but an interesting note. Suppose we denote $H(n) = 1/1 + 1/2 + ... + 1/n$ then $H(n!) - H((n-1)!) approx log(n)$ for large n. Does this give a hint? ;)
    – Roupam Ghosh
    Jul 11 '11 at 4:14






  • 5




    Here is a weakly related question: What is a textbook, or even a popularization for the general public, that (1) discusses infinite series, but (2) does not have an explanation for the divergence of this exact series?
    – GEdgar
    Nov 3 '13 at 19:50










  • to avoid defining the logarithm, use the Cauchy condensation test to show that $sum 1/n$ converges iff $sum 1$ converges
    – reuns
    Jan 30 '16 at 23:31












  • These are two of my favourite papers: The Harmonic Series Diverges Again and Again and More Proofs of Divergence of the Harmonic Series. See these.
    – user249332
    Mar 9 '16 at 22:56










  • If it converges, then it contradicts the dominated convergence theorem. This proof is easily comprehensible if you know the dominated convergence theorem, but that theorem is not the most comprehensible.
    – Oiler
    Oct 6 '16 at 23:08















up vote
120
down vote

favorite
54









up vote
120
down vote

favorite
54






54





Can someone give a simple explanation as to why the harmonic series




$$sum_{n=1}^inftyfrac1n=frac 1 1 + frac 12 + frac 13 + cdots $$




doesn't converge, on the other hand it grows very slowly?



I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.










share|cite|improve this question















Can someone give a simple explanation as to why the harmonic series




$$sum_{n=1}^inftyfrac1n=frac 1 1 + frac 12 + frac 13 + cdots $$




doesn't converge, on the other hand it grows very slowly?



I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.







calculus sequences-and-series harmonic-numbers






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edited Jul 27 '16 at 15:16









haqnatural

20.6k72457




20.6k72457










asked Jul 21 '10 at 5:00









bryn

3,43773130




3,43773130








  • 3




    This is not meant to be an answer but an interesting note. Suppose we denote $H(n) = 1/1 + 1/2 + ... + 1/n$ then $H(n!) - H((n-1)!) approx log(n)$ for large n. Does this give a hint? ;)
    – Roupam Ghosh
    Jul 11 '11 at 4:14






  • 5




    Here is a weakly related question: What is a textbook, or even a popularization for the general public, that (1) discusses infinite series, but (2) does not have an explanation for the divergence of this exact series?
    – GEdgar
    Nov 3 '13 at 19:50










  • to avoid defining the logarithm, use the Cauchy condensation test to show that $sum 1/n$ converges iff $sum 1$ converges
    – reuns
    Jan 30 '16 at 23:31












  • These are two of my favourite papers: The Harmonic Series Diverges Again and Again and More Proofs of Divergence of the Harmonic Series. See these.
    – user249332
    Mar 9 '16 at 22:56










  • If it converges, then it contradicts the dominated convergence theorem. This proof is easily comprehensible if you know the dominated convergence theorem, but that theorem is not the most comprehensible.
    – Oiler
    Oct 6 '16 at 23:08
















  • 3




    This is not meant to be an answer but an interesting note. Suppose we denote $H(n) = 1/1 + 1/2 + ... + 1/n$ then $H(n!) - H((n-1)!) approx log(n)$ for large n. Does this give a hint? ;)
    – Roupam Ghosh
    Jul 11 '11 at 4:14






  • 5




    Here is a weakly related question: What is a textbook, or even a popularization for the general public, that (1) discusses infinite series, but (2) does not have an explanation for the divergence of this exact series?
    – GEdgar
    Nov 3 '13 at 19:50










  • to avoid defining the logarithm, use the Cauchy condensation test to show that $sum 1/n$ converges iff $sum 1$ converges
    – reuns
    Jan 30 '16 at 23:31












  • These are two of my favourite papers: The Harmonic Series Diverges Again and Again and More Proofs of Divergence of the Harmonic Series. See these.
    – user249332
    Mar 9 '16 at 22:56










  • If it converges, then it contradicts the dominated convergence theorem. This proof is easily comprehensible if you know the dominated convergence theorem, but that theorem is not the most comprehensible.
    – Oiler
    Oct 6 '16 at 23:08










3




3




This is not meant to be an answer but an interesting note. Suppose we denote $H(n) = 1/1 + 1/2 + ... + 1/n$ then $H(n!) - H((n-1)!) approx log(n)$ for large n. Does this give a hint? ;)
– Roupam Ghosh
Jul 11 '11 at 4:14




This is not meant to be an answer but an interesting note. Suppose we denote $H(n) = 1/1 + 1/2 + ... + 1/n$ then $H(n!) - H((n-1)!) approx log(n)$ for large n. Does this give a hint? ;)
– Roupam Ghosh
Jul 11 '11 at 4:14




5




5




Here is a weakly related question: What is a textbook, or even a popularization for the general public, that (1) discusses infinite series, but (2) does not have an explanation for the divergence of this exact series?
– GEdgar
Nov 3 '13 at 19:50




Here is a weakly related question: What is a textbook, or even a popularization for the general public, that (1) discusses infinite series, but (2) does not have an explanation for the divergence of this exact series?
– GEdgar
Nov 3 '13 at 19:50












to avoid defining the logarithm, use the Cauchy condensation test to show that $sum 1/n$ converges iff $sum 1$ converges
– reuns
Jan 30 '16 at 23:31






to avoid defining the logarithm, use the Cauchy condensation test to show that $sum 1/n$ converges iff $sum 1$ converges
– reuns
Jan 30 '16 at 23:31














These are two of my favourite papers: The Harmonic Series Diverges Again and Again and More Proofs of Divergence of the Harmonic Series. See these.
– user249332
Mar 9 '16 at 22:56




These are two of my favourite papers: The Harmonic Series Diverges Again and Again and More Proofs of Divergence of the Harmonic Series. See these.
– user249332
Mar 9 '16 at 22:56












If it converges, then it contradicts the dominated convergence theorem. This proof is easily comprehensible if you know the dominated convergence theorem, but that theorem is not the most comprehensible.
– Oiler
Oct 6 '16 at 23:08






If it converges, then it contradicts the dominated convergence theorem. This proof is easily comprehensible if you know the dominated convergence theorem, but that theorem is not the most comprehensible.
– Oiler
Oct 6 '16 at 23:08












21 Answers
21






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up vote
134
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accepted










Let's group the terms as follows:



Group $1$ : $displaystylefrac11qquad$ ($1$ term)



Group $2$ : $displaystylefrac12+frac13qquad$($2$ terms)



Group $3$ : $displaystylefrac14+frac15+frac16+frac17qquad$($4$ terms)



Group $4$ : $displaystylefrac18+frac19+cdots+frac1{15}qquad$ ($8$ terms)



$quadvdots$



In general, group $n$ contains $2^{n-1}$ terms. But also, notice that the smallest element in group $n$ is larger than $dfrac1{2^n}$. For example all elements in group $2$ are larger than $dfrac1{2^2}$. So the sum of the terms in each group is larger than $2^{n-1} cdot dfrac1{2^n} = dfrac1{2}$. Since there are infinitely many groups, and the sum in each group is larger than $dfrac1{2}$, it follows that the total sum is infinite.



This proof is often attributed to Nicole Oresme.






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  • 8




    +1: This is nice: it's easy to turn this into a rigorous proof, and it even gives you a lower bound for the order of growth!
    – Simon Nickerson
    Jul 21 '10 at 5:19






  • 2




    I assume you mean that group 4 as 8 terms? Or do you mean to go all the way to 1/23?
    – Tomas Aschan
    Jul 21 '10 at 7:37






  • 1




    Is there a closed-form function for this value?
    – John Gietzen
    Jul 21 '10 at 18:29






  • 2




    Interestingly, this proof goes as far back as Nicole Oresme in the 14th century. Wikipedia has a nice display of this proof [en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]
    – Neil Mayhew
    Jul 22 '10 at 13:20






  • 1




    @John: There's no explicit closed-form, but they're generally known as the Harmonic Numbers; there are a number of identities involving them (how to sum them or sum multiples of them, etc.)
    – Steven Stadnicki
    Jul 10 '11 at 21:23


















up vote
37
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There is a fantastic collection of $20$ different proofs that this series diverges. I recommend you read it (it can be found here). I especially like proof $14$, which appeals to triangular numbers for a sort of cameo role.





EDIT



It seems the original link is broken, due to the author moving to his own site. So I followed up and found the new link. In addition, the author has an extended addendum, bringing the total number of proofs to 42+.






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  • 2




    Proof 6 is also nice.
    – leonbloy
    Mar 15 '13 at 20:51










  • Apparently, the list has been updated.
    – David Mitra
    Jun 19 '13 at 16:15


















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23
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Let's group the terms as follows:$$A=frac11+frac12+frac13+frac14+cdots\ $$
$$
A=underbrace{(frac{1}{1}+frac{1}{2}+frac{1}{3}+cdots+frac{1}{9})}_{color{red} {9- terms}}
+underbrace{(frac{1}{10}+frac{1}{11}+frac{1}{12}+cdots+frac{1}{99})}_{color{red} {90- terms}}\+underbrace{(frac{1}{101}+frac{1}{102}+frac{1}{103}+cdots+frac{1}{999})}_{color{red} {900- terms}}+cdots \ to $$
$$\A>9 times(frac{1}{10})+(99-10+1)times frac{1}{100}+(999-100+1)times frac{1}{1000}+... \A>frac{9}{10}+frac{90}{100}+frac{90}{100}+frac{900}{1000}+...\ to A>underbrace{frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+...}_{color{red} {text{ m group} ,text{ and} space mto infty}} to infty
$$



Showing that $A$ diverges by grouping numbers.






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    up vote
    22
    down vote













    The answer given by AgCl is a classic one. And possibly pedagogically best; I don't know.



    I also like the following argument. I'm not sure what students who are new to the topic will think about it.



    Suppose 1 + 1/2 + 1/3 + 1/4 + ... adds up to some finite total S. Now group terms in the following way:



    $$1 + frac{1}{2} > frac{1}{2} + frac{1}{2} = frac{2}{2} = 1$$



    $$frac{1}{3} + frac{1}{4} > frac{1}{4} + frac{1}{4} = frac{2}{4} = frac{1}{2}$$



    $$frac{1}{5} + frac{1}{6} > frac{1}{6} + frac{1}{6} = frac{2}{6} = frac{1}{3}$$



    Continuing in this way, we get $S > S$, a contradiction.






    share|cite|improve this answer



















    • 1




      Not really. From $S_n > T_n$ you can only conclude that $lim S_n ge lim T_n$.
      – lhf
      Jul 10 '11 at 21:24








    • 6




      @lhf: That's right, but that can be easily fixed here (with $S_n = 1 + 1/2 + dots + 1/2n$ and $T_n = 1 + 1/2 + dots + 1/n$): we can use a better inequality, like say $S_n ge T_n + 1/2$ (using just the first step) to conclude that $lim S_n ge lim T_n + 1/2$, contradicting $S = lim S_n = lim T_n$.
      – ShreevatsaR
      Jul 11 '11 at 4:18


















    up vote
    19
    down vote













    An alternative proof (translated and adapted from this comment by Filipe Oliveira, in Portuguese, posted also here). Let $ f(x)=ln(1+x)$. Then $f'(x)=dfrac {1}{1+x}$ and $ f'(0)=1$. Hence



    $$displaystylelim_{xto 0}dfrac{ln(1+x)}{x}=lim_{xto 0}dfrac{ln(1+x)-ln(1)}{x-0}=1,$$



    and



    $$ displaystylelim_{ntoinfty} dfrac{lnleft(1+dfrac{1}{n}right)}{dfrac {1}{n}}=1>0.$$



    So, the series $displaystylesumdfrac{1}{n}$ and $displaystylesumlnleft(1+dfrac {1}{n}right)$ are both convergent or divergent. Since



    $$lnleft(1+dfrac {1}{n}right)=lnleft(dfrac{n+1}{n}right)=ln (n+1)-ln(n),$$



    we have



    $$displaystylesum_{n=1}^Nlnleft(1+dfrac {1}{n}right)=ln(N+1)-ln(1)=ln(N+1).$$



    Thus $displaystylesum_{n=1}^{infty}lnleft(1+dfrac {1}{n}right)$ is divergent and so is $displaystylesum_{n=1}^{infty}dfrac{1}{n}$.






    share|cite|improve this answer




























      up vote
      19
      down vote













      This is not as good an answer as AgCl's, nonetheless people may find it interesting.



      If you're used to calculus then you might notice that the sum $$
      1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n}$$ is very close to the integral from $1$ to $n$ of $frac{1}{x}$. This definite integral is ln(n), so you should expect $1+frac{1}{2}+frac{1}{3}+dots+
      frac{1}{n}$ to grow like $ln(n)$.



      Although this argument can be made rigorous, it's still unsatisfying because it depends on the fact that the derivative of $ln(x)$ is $frac{1}{x}$, which is probably harder than the original question. Nonetheless it does illustrate a good general heuristic for quickly determining how sums behave if you already know calculus.






      share|cite|improve this answer



















      • 2




        If you look at a Riemann sum for intervals with width 1, you can pretty quickly see that the integral of 1/x from 1 to infinity must be less than the sum of the harmonic series.
        – Isaac
        Jul 21 '10 at 5:51










      • Thank you for adding this answer. I was hoping to avoid an answer that involved integration, so I also prefer AgCl's answer. But I am happy to see more than one demonstration/proof.
        – bryn
        Jul 22 '10 at 11:33










      • The sum is closer to the integral from $frac{1}{2}$ to $n+frac{1}{2}$ of $frac{1}{x}$, which is $log(2n+1)$ math.stackexchange.com/a/1602945/134791
        – Jaume Oliver Lafont
        Jan 25 '16 at 23:02


















      up vote
      17
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      This was bumped, so I'll add a proof sweet proof I saw in this site. Exponentiate $H_n$ and get $$e^{H_n}=prod_{k=1}^n e^{1/k}gtprod_{k=1}^nleft(1+frac{1}{k}right)=n+1.$$ Therefore, $H_ngtlog(n+1)$, so we are done. We used $e^xgt1+x$ and telescoped the resulting product.






      share|cite|improve this answer





















      • Oh, that's unique.
        – Simply Beautiful Art
        Oct 7 '16 at 1:12


















      up vote
      13
      down vote













      Another interesting proof is based upon one of the consequences of the Lagrange's theorem applied on $ln(x)$ function, namely:



      $$frac{1}{k+1} < ln(k+1)-ln(k)<frac{1}{k} space , space kinmathbb{N} ,space k>0$$



      Taking $k=1,2,...,n$ values to the inequality and then summing all relations, we get the required result.



      The proof is complete.






      share|cite|improve this answer






























        up vote
        11
        down vote













        There also exists a proof for the divergence of the harmonic series that involves the Integral Test. It goes as follows.



        It is possible to prove that the harmonic series diverges by comparing its sum with an improper integral. Specifically, consider the arrangement of rectangles shown in the figure of $ y = dfrac {1}{x} $:



        y=1/x



        Each rectangle is $1$ unit wide and $frac{1}{n}$ units high, so the total area of the rectangles is the sum of the harmonic series: $$ displaystylesum left( text {enclosed rectangle are} right) = displaystylesum_{k=1}^{infty} dfrac {1}{k}. $$Now, the total area under the curve is given by $$ displaystyleint_{1}^{infty} dfrac {mathrm{d}x}{x} = infty. $$Since this area is entirely contained within the rectangles, the total area of the rectangles must be infinite as well. More precisely, this proves that $$ displaystylesum_{n=1}^{k} dfrac {1}{n} > displaystyleint_{1}^{k+1} dfrac {mathrm{d}x}{x} = ln (k+1). $$This is the backbone of what we know today as the integral test.



        Interestingly, the alternating harmonic series does converge: $$ displaystylesum_{n=1}^{infty} dfrac {(-1)^n}{n} = ln 2. $$And so does the $p$-harmonic series with $p>1$.






        share|cite|improve this answer






























          up vote
          8
          down vote













          Let's assume that $sum_{n=1}^{infty}frac1n=:Hin mathbb{R}$, then
          $$H=frac11+frac12+frac13+frac14+frac15+frac16 +ldots $$
          $$Hgeqslant frac11+frac12 +frac14+frac14+frac16+frac16+ldots$$
          $$Hgeqslant frac11+frac12+frac12+frac13+frac14+frac15+ldots$$
          $$Hgeqslant frac12 +H Rightarrow 0geqslant frac12$$
          Since the last inequality doesn't hold, we can conclude that the sum doesn't converge.






          share|cite|improve this answer




























            up vote
            8
            down vote













            $$int_{0}^{infty}e^{-nx}dx=frac1n$$



            $$sum_{n=1}^{infty}int_{0}^{infty}e^{-nx}dx=lim_{ m to infty}sum_{n=1}^{m}frac1n$$



            using the law of Geometric series



            $$int_{0}^{infty}(frac{1}{1-e^{-x}}-1)dx=lim_{ m to infty}H_m$$



            $$lim_{ m to infty}H_m=left [ ln(e^x-1)-x right ]_0^{infty}toinfty$$






            share|cite|improve this answer























            • Hm, the lower bound goes to $-infty$ it appears.
              – Simply Beautiful Art
              Oct 6 '16 at 16:40










            • @SimpleArt The upper bound goes to 0 and The Lower goes to $+infty$ ,,,$-ln 0^+$
              – mhd.math
              Oct 7 '16 at 11:20












            • Oh right, duh, didn't quite use that FTOC correctly.
              – Simply Beautiful Art
              Oct 7 '16 at 13:19


















            up vote
            7
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            Another (different) answer, by the Cauchy Condensation Test :



            $$sum_{n=1}^infty frac{1}{n} < infty iff sum_{n=1}^infty 2^n frac{1}{2^n} = sum_{n=1}^infty 1< infty $$



            The latter is obviously divergent, therefore the former diverges. This is THE shortest proof there is.






            share|cite|improve this answer




























              up vote
              7
              down vote













              Suppose to the contrary that converges.



              Let $s_n$ denote the $n$-th partial sum. Since the serie converges, $(s_n)$ is a Cauchy sequence. Let $varepsilon = 1/3$, then there is some $n_0$ such that $|s_q-s_p|< 1/3$ for all $q>pge n_0$. Let $q=2n_0$ and $p=n_0$. Then



              $$frac{1}{3}>bigg|sum_{n=n_0+1}^{2n_0} frac{1}{n}bigg|gebigg|sum_{n=n_0+1}^{2n_0} frac{1}{2n_0}bigg|=frac{1}{2}$$



              a contradiction. Then this contradiction shows that the series diverges.






              share|cite|improve this answer




























                up vote
                7
                down vote













                enter image description here



                First suppose $displaystyle A=frac11+frac12+frac13+frac14+cdots$ converges
                then show that $A>A$. That's paradox.






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                • 11




                  Ideally use Latex.
                  – Meow
                  Nov 1 '13 at 17:13








                • 2




                  I am afraid that this approach is incorrect, since similar versions of it can be applied to convergent infinite series. The reason that it does not work is because the number of terms is infinite. All you've proven is that the second series approaches the final value faster than the first one. But whether this value is ultimately finite or not, you have not shown.
                  – Lucian
                  Jan 14 '15 at 16:32




















                up vote
                6
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                A non-rigorous explanation I thought of once: consider a savings scheme where you put a dollar in your piggy bank every day. So after $n$ days, you have $n$ dollars; clearly, your savings approach infinity. On the other hand, each day you add an additional $1/n$ proportion of your existing savings, "so" (the non-rigorous step) the accumulated percentage after $n$ days is $1 + 1/2 + cdots + 1/n$.



                This can be made rigorous through the infinite product argument
                $$prod_{n = 1}^infty (1 + tfrac{1}{n}) < infty iff sum_{n = 1}^infty frac{1}{n} < infty$$
                which is obtained, essentially, by taking the logarithm of the left-hand side and using the power series for $log (1 + x)$.






                share|cite|improve this answer




























                  up vote
                  6
                  down vote













                  Another answer that's very similar to others. But it's prettier, and perhaps easier to understand for the 9-th grade student who asked the same question here.



                  The student's question was ... does the sum equal some number $S$. But, look:



                  enter image description here



                  So, whatever it is, $S$ is larger than the sum of the infinite string of $tfrac12$'s shown in the last line. No number can be this large, so $S$ can't be equal to any number. Mathematicians say that the series "diverges to infinity".






                  share|cite|improve this answer






























                    up vote
                    5
                    down vote













                    I think the integral test gives the most intuitive explanation. Observe that $$int^n_1 frac1x dx= log n$$ The sum $displaystylesum^n_{k=1}frac1k$ can be viewed as the area of $n$ rectangles of height $frac1k$, width $1$ (with the first one having it's left hand side on the y axis, and all having their bottom on the x axis). The graph of $xmapsto frac1x$ can be drawn under these, so the sum will grow with $n$ at (least) as fast as the integral - hence will grow (at least) logarithmically.






                    share|cite|improve this answer























                    • use log to get nice formatting for $log$
                      – Tyler
                      Nov 1 '13 at 17:24


















                    up vote
                    4
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                    Let be the partial sum $H_n = frac11 + frac12 + frac13 + cdots + frac1n$. Using Cesàro-Stolz:
                    $$
                    lim_{ntoinfty}frac{H_n}{log n} = lim_{ntoinfty}frac{H_{n+1}-H_n}{log(n+1)-log n} = lim_{ntoinfty}frac{frac1{n+1}}{log(1+1/n)}
                    = lim_{ntoinfty}frac{frac1{n+1}}{frac1n} = 1
                    $$
                    and
                    $$sum_{n=1}^inftyfrac1n = lim_{ntoinfty}H_n = infty.$$






                    share|cite|improve this answer




























                      up vote
                      2
                      down vote













                      We all know that $$sum_{n=1}^inftyfrac1n=frac 1 1 + frac 12 + frac 13 + cdots $$ diverges and grows very slowly!! I have seen many proofs of the result but recently found the following: $$S =frac 1 1 + frac 12 + frac 13 +frac 14+ frac 15+ frac 16+ cdots$$ $$> frac 12+frac 12+ frac 14+ frac 14+ frac 16+ frac 16+ cdots =frac 1 1 + frac 12 + frac 13 +cdots = S.$$
                      In this way we see that $S > S$.






                      share|cite|improve this answer





















                      • O.o This. Is. Amazing!! =)
                        – user378947
                        Dec 12 '16 at 1:18










                      • You can also see it here math.stackexchange.com/questions/1160527/…
                        – user8795
                        Dec 12 '16 at 1:20










                      • I have saved this to my personal The Book :) That being said... Come on! The last inequality itself is proof enough!! :P
                        – user378947
                        Dec 12 '16 at 1:31


















                      up vote
                      2
                      down vote













                      Using Euler's form of the Harmonic numbers,



                      $$sum_{k=1}^nfrac1k=int_0^1frac{1-x^n}{1-x}dx$$



                      $$begin{align}
                      lim_{ntoinfty}sum_{k=1}^nfrac1k & =lim_{ntoinfty}int_0^1frac{1-x^n}{1-x}dx \
                      & =int_0^1frac1{1-x}dx \
                      & =left.lim_{pto1^+}-ln(1-x)right]_0^p \
                      & to+infty
                      end{align}$$





                      Using the Taylor expansion of $ln(1-x)$,



                      $$-ln(1-x)=x+frac{x^2}2+frac{x^3}3+frac{x^4}4+dots$$



                      $$-ln(1-1)=1+frac12+frac13+frac14+dotsquad $$





                      Using Euler's relationship between the Riemann zeta function and the Dirichlet eta function,



                      $$begin{align}
                      sum_{k=1}^inftyfrac1{k^s} & =frac1{1-2^{1-s}}sum_{k=1}^inftyfrac{(-1)^{k+1}}{k^s} \
                      sum_{k=1}^inftyfrac1k & =frac10sum_{k=1}^inftyfrac{(-1)^{k+1}}ktag{$s=1$} \
                      & to+infty
                      end{align}$$






                      share|cite|improve this answer























                      • But isn't the series for $ln(1-x)$ only valid for $-1le x<1$?
                        – TheSimpliFire
                        Mar 17 at 9:31












                      • Yes, but since the limit as $xto1$ in $x^n$ is monotone, it equals the asked series, if they exist.
                        – Simply Beautiful Art
                        Mar 19 at 2:33


















                      up vote
                      -3
                      down vote













                      A series converges if and only if the tail of the series tends to zero, i.e. the summation from N to infinity tends to zero for N to infinity. But in case of the Harmonic series, we have that the summation from N to 2 N is larger than the smallest term ( which is 1/(2N)), times the number of terms N, which yields 1/2. So, the tail clearly does not tend to zero for N to infinity.






                      share|cite|improve this answer



















                      • 2




                        What do you think you were adding that hasn't been addressed thoroughly?
                        – user223391
                        Jan 31 '16 at 1:05










                      • @avid19 I added an argument that's tl;dr-proof.
                        – Count Iblis
                        Jan 31 '16 at 4:29










                      • @ZacharySelk My proof is the best, as it's the most concise, self contained proof given. Unlike the other proofs my proof can be modified into an argument that a 6 year old could understand (with some effort).
                        – Count Iblis
                        Oct 6 '16 at 22:35






                      • 2




                        @CountIblis Your proof is mathematically identical to the six-year-earlier (and top-voted, and accepted) answer by AgCl, which also explains the situation much more clearly. (Note that bringing tails into the picture is unnecessary, and just adds a layer of complexity: we can reason about the blocks directly as AgCl does, and as you do essentially, to show that the harmonic series goes to infinity.)
                        – Noah Schweber
                        Oct 11 '16 at 1:52













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                      21 Answers
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                      21 Answers
                      21






                      active

                      oldest

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                      active

                      oldest

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                      active

                      oldest

                      votes








                      up vote
                      134
                      down vote



                      accepted










                      Let's group the terms as follows:



                      Group $1$ : $displaystylefrac11qquad$ ($1$ term)



                      Group $2$ : $displaystylefrac12+frac13qquad$($2$ terms)



                      Group $3$ : $displaystylefrac14+frac15+frac16+frac17qquad$($4$ terms)



                      Group $4$ : $displaystylefrac18+frac19+cdots+frac1{15}qquad$ ($8$ terms)



                      $quadvdots$



                      In general, group $n$ contains $2^{n-1}$ terms. But also, notice that the smallest element in group $n$ is larger than $dfrac1{2^n}$. For example all elements in group $2$ are larger than $dfrac1{2^2}$. So the sum of the terms in each group is larger than $2^{n-1} cdot dfrac1{2^n} = dfrac1{2}$. Since there are infinitely many groups, and the sum in each group is larger than $dfrac1{2}$, it follows that the total sum is infinite.



                      This proof is often attributed to Nicole Oresme.






                      share|cite|improve this answer



















                      • 8




                        +1: This is nice: it's easy to turn this into a rigorous proof, and it even gives you a lower bound for the order of growth!
                        – Simon Nickerson
                        Jul 21 '10 at 5:19






                      • 2




                        I assume you mean that group 4 as 8 terms? Or do you mean to go all the way to 1/23?
                        – Tomas Aschan
                        Jul 21 '10 at 7:37






                      • 1




                        Is there a closed-form function for this value?
                        – John Gietzen
                        Jul 21 '10 at 18:29






                      • 2




                        Interestingly, this proof goes as far back as Nicole Oresme in the 14th century. Wikipedia has a nice display of this proof [en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]
                        – Neil Mayhew
                        Jul 22 '10 at 13:20






                      • 1




                        @John: There's no explicit closed-form, but they're generally known as the Harmonic Numbers; there are a number of identities involving them (how to sum them or sum multiples of them, etc.)
                        – Steven Stadnicki
                        Jul 10 '11 at 21:23















                      up vote
                      134
                      down vote



                      accepted










                      Let's group the terms as follows:



                      Group $1$ : $displaystylefrac11qquad$ ($1$ term)



                      Group $2$ : $displaystylefrac12+frac13qquad$($2$ terms)



                      Group $3$ : $displaystylefrac14+frac15+frac16+frac17qquad$($4$ terms)



                      Group $4$ : $displaystylefrac18+frac19+cdots+frac1{15}qquad$ ($8$ terms)



                      $quadvdots$



                      In general, group $n$ contains $2^{n-1}$ terms. But also, notice that the smallest element in group $n$ is larger than $dfrac1{2^n}$. For example all elements in group $2$ are larger than $dfrac1{2^2}$. So the sum of the terms in each group is larger than $2^{n-1} cdot dfrac1{2^n} = dfrac1{2}$. Since there are infinitely many groups, and the sum in each group is larger than $dfrac1{2}$, it follows that the total sum is infinite.



                      This proof is often attributed to Nicole Oresme.






                      share|cite|improve this answer



















                      • 8




                        +1: This is nice: it's easy to turn this into a rigorous proof, and it even gives you a lower bound for the order of growth!
                        – Simon Nickerson
                        Jul 21 '10 at 5:19






                      • 2




                        I assume you mean that group 4 as 8 terms? Or do you mean to go all the way to 1/23?
                        – Tomas Aschan
                        Jul 21 '10 at 7:37






                      • 1




                        Is there a closed-form function for this value?
                        – John Gietzen
                        Jul 21 '10 at 18:29






                      • 2




                        Interestingly, this proof goes as far back as Nicole Oresme in the 14th century. Wikipedia has a nice display of this proof [en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]
                        – Neil Mayhew
                        Jul 22 '10 at 13:20






                      • 1




                        @John: There's no explicit closed-form, but they're generally known as the Harmonic Numbers; there are a number of identities involving them (how to sum them or sum multiples of them, etc.)
                        – Steven Stadnicki
                        Jul 10 '11 at 21:23













                      up vote
                      134
                      down vote



                      accepted







                      up vote
                      134
                      down vote



                      accepted






                      Let's group the terms as follows:



                      Group $1$ : $displaystylefrac11qquad$ ($1$ term)



                      Group $2$ : $displaystylefrac12+frac13qquad$($2$ terms)



                      Group $3$ : $displaystylefrac14+frac15+frac16+frac17qquad$($4$ terms)



                      Group $4$ : $displaystylefrac18+frac19+cdots+frac1{15}qquad$ ($8$ terms)



                      $quadvdots$



                      In general, group $n$ contains $2^{n-1}$ terms. But also, notice that the smallest element in group $n$ is larger than $dfrac1{2^n}$. For example all elements in group $2$ are larger than $dfrac1{2^2}$. So the sum of the terms in each group is larger than $2^{n-1} cdot dfrac1{2^n} = dfrac1{2}$. Since there are infinitely many groups, and the sum in each group is larger than $dfrac1{2}$, it follows that the total sum is infinite.



                      This proof is often attributed to Nicole Oresme.






                      share|cite|improve this answer














                      Let's group the terms as follows:



                      Group $1$ : $displaystylefrac11qquad$ ($1$ term)



                      Group $2$ : $displaystylefrac12+frac13qquad$($2$ terms)



                      Group $3$ : $displaystylefrac14+frac15+frac16+frac17qquad$($4$ terms)



                      Group $4$ : $displaystylefrac18+frac19+cdots+frac1{15}qquad$ ($8$ terms)



                      $quadvdots$



                      In general, group $n$ contains $2^{n-1}$ terms. But also, notice that the smallest element in group $n$ is larger than $dfrac1{2^n}$. For example all elements in group $2$ are larger than $dfrac1{2^2}$. So the sum of the terms in each group is larger than $2^{n-1} cdot dfrac1{2^n} = dfrac1{2}$. Since there are infinitely many groups, and the sum in each group is larger than $dfrac1{2}$, it follows that the total sum is infinite.



                      This proof is often attributed to Nicole Oresme.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jul 29 '14 at 5:35









                      Tunk-Fey

                      22.9k868100




                      22.9k868100










                      answered Jul 21 '10 at 5:13









                      AgCl

                      3,90753134




                      3,90753134








                      • 8




                        +1: This is nice: it's easy to turn this into a rigorous proof, and it even gives you a lower bound for the order of growth!
                        – Simon Nickerson
                        Jul 21 '10 at 5:19






                      • 2




                        I assume you mean that group 4 as 8 terms? Or do you mean to go all the way to 1/23?
                        – Tomas Aschan
                        Jul 21 '10 at 7:37






                      • 1




                        Is there a closed-form function for this value?
                        – John Gietzen
                        Jul 21 '10 at 18:29






                      • 2




                        Interestingly, this proof goes as far back as Nicole Oresme in the 14th century. Wikipedia has a nice display of this proof [en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]
                        – Neil Mayhew
                        Jul 22 '10 at 13:20






                      • 1




                        @John: There's no explicit closed-form, but they're generally known as the Harmonic Numbers; there are a number of identities involving them (how to sum them or sum multiples of them, etc.)
                        – Steven Stadnicki
                        Jul 10 '11 at 21:23














                      • 8




                        +1: This is nice: it's easy to turn this into a rigorous proof, and it even gives you a lower bound for the order of growth!
                        – Simon Nickerson
                        Jul 21 '10 at 5:19






                      • 2




                        I assume you mean that group 4 as 8 terms? Or do you mean to go all the way to 1/23?
                        – Tomas Aschan
                        Jul 21 '10 at 7:37






                      • 1




                        Is there a closed-form function for this value?
                        – John Gietzen
                        Jul 21 '10 at 18:29






                      • 2




                        Interestingly, this proof goes as far back as Nicole Oresme in the 14th century. Wikipedia has a nice display of this proof [en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]
                        – Neil Mayhew
                        Jul 22 '10 at 13:20






                      • 1




                        @John: There's no explicit closed-form, but they're generally known as the Harmonic Numbers; there are a number of identities involving them (how to sum them or sum multiples of them, etc.)
                        – Steven Stadnicki
                        Jul 10 '11 at 21:23








                      8




                      8




                      +1: This is nice: it's easy to turn this into a rigorous proof, and it even gives you a lower bound for the order of growth!
                      – Simon Nickerson
                      Jul 21 '10 at 5:19




                      +1: This is nice: it's easy to turn this into a rigorous proof, and it even gives you a lower bound for the order of growth!
                      – Simon Nickerson
                      Jul 21 '10 at 5:19




                      2




                      2




                      I assume you mean that group 4 as 8 terms? Or do you mean to go all the way to 1/23?
                      – Tomas Aschan
                      Jul 21 '10 at 7:37




                      I assume you mean that group 4 as 8 terms? Or do you mean to go all the way to 1/23?
                      – Tomas Aschan
                      Jul 21 '10 at 7:37




                      1




                      1




                      Is there a closed-form function for this value?
                      – John Gietzen
                      Jul 21 '10 at 18:29




                      Is there a closed-form function for this value?
                      – John Gietzen
                      Jul 21 '10 at 18:29




                      2




                      2




                      Interestingly, this proof goes as far back as Nicole Oresme in the 14th century. Wikipedia has a nice display of this proof [en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]
                      – Neil Mayhew
                      Jul 22 '10 at 13:20




                      Interestingly, this proof goes as far back as Nicole Oresme in the 14th century. Wikipedia has a nice display of this proof [en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29]
                      – Neil Mayhew
                      Jul 22 '10 at 13:20




                      1




                      1




                      @John: There's no explicit closed-form, but they're generally known as the Harmonic Numbers; there are a number of identities involving them (how to sum them or sum multiples of them, etc.)
                      – Steven Stadnicki
                      Jul 10 '11 at 21:23




                      @John: There's no explicit closed-form, but they're generally known as the Harmonic Numbers; there are a number of identities involving them (how to sum them or sum multiples of them, etc.)
                      – Steven Stadnicki
                      Jul 10 '11 at 21:23










                      up vote
                      37
                      down vote













                      There is a fantastic collection of $20$ different proofs that this series diverges. I recommend you read it (it can be found here). I especially like proof $14$, which appeals to triangular numbers for a sort of cameo role.





                      EDIT



                      It seems the original link is broken, due to the author moving to his own site. So I followed up and found the new link. In addition, the author has an extended addendum, bringing the total number of proofs to 42+.






                      share|cite|improve this answer



















                      • 2




                        Proof 6 is also nice.
                        – leonbloy
                        Mar 15 '13 at 20:51










                      • Apparently, the list has been updated.
                        – David Mitra
                        Jun 19 '13 at 16:15















                      up vote
                      37
                      down vote













                      There is a fantastic collection of $20$ different proofs that this series diverges. I recommend you read it (it can be found here). I especially like proof $14$, which appeals to triangular numbers for a sort of cameo role.





                      EDIT



                      It seems the original link is broken, due to the author moving to his own site. So I followed up and found the new link. In addition, the author has an extended addendum, bringing the total number of proofs to 42+.






                      share|cite|improve this answer



















                      • 2




                        Proof 6 is also nice.
                        – leonbloy
                        Mar 15 '13 at 20:51










                      • Apparently, the list has been updated.
                        – David Mitra
                        Jun 19 '13 at 16:15













                      up vote
                      37
                      down vote










                      up vote
                      37
                      down vote









                      There is a fantastic collection of $20$ different proofs that this series diverges. I recommend you read it (it can be found here). I especially like proof $14$, which appeals to triangular numbers for a sort of cameo role.





                      EDIT



                      It seems the original link is broken, due to the author moving to his own site. So I followed up and found the new link. In addition, the author has an extended addendum, bringing the total number of proofs to 42+.






                      share|cite|improve this answer














                      There is a fantastic collection of $20$ different proofs that this series diverges. I recommend you read it (it can be found here). I especially like proof $14$, which appeals to triangular numbers for a sort of cameo role.





                      EDIT



                      It seems the original link is broken, due to the author moving to his own site. So I followed up and found the new link. In addition, the author has an extended addendum, bringing the total number of proofs to 42+.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 3 '13 at 18:26

























                      answered Jul 11 '11 at 4:08









                      davidlowryduda

                      73.8k7116248




                      73.8k7116248








                      • 2




                        Proof 6 is also nice.
                        – leonbloy
                        Mar 15 '13 at 20:51










                      • Apparently, the list has been updated.
                        – David Mitra
                        Jun 19 '13 at 16:15














                      • 2




                        Proof 6 is also nice.
                        – leonbloy
                        Mar 15 '13 at 20:51










                      • Apparently, the list has been updated.
                        – David Mitra
                        Jun 19 '13 at 16:15








                      2




                      2




                      Proof 6 is also nice.
                      – leonbloy
                      Mar 15 '13 at 20:51




                      Proof 6 is also nice.
                      – leonbloy
                      Mar 15 '13 at 20:51












                      Apparently, the list has been updated.
                      – David Mitra
                      Jun 19 '13 at 16:15




                      Apparently, the list has been updated.
                      – David Mitra
                      Jun 19 '13 at 16:15










                      up vote
                      23
                      down vote













                      Let's group the terms as follows:$$A=frac11+frac12+frac13+frac14+cdots\ $$
                      $$
                      A=underbrace{(frac{1}{1}+frac{1}{2}+frac{1}{3}+cdots+frac{1}{9})}_{color{red} {9- terms}}
                      +underbrace{(frac{1}{10}+frac{1}{11}+frac{1}{12}+cdots+frac{1}{99})}_{color{red} {90- terms}}\+underbrace{(frac{1}{101}+frac{1}{102}+frac{1}{103}+cdots+frac{1}{999})}_{color{red} {900- terms}}+cdots \ to $$
                      $$\A>9 times(frac{1}{10})+(99-10+1)times frac{1}{100}+(999-100+1)times frac{1}{1000}+... \A>frac{9}{10}+frac{90}{100}+frac{90}{100}+frac{900}{1000}+...\ to A>underbrace{frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+...}_{color{red} {text{ m group} ,text{ and} space mto infty}} to infty
                      $$



                      Showing that $A$ diverges by grouping numbers.






                      share|cite|improve this answer



























                        up vote
                        23
                        down vote













                        Let's group the terms as follows:$$A=frac11+frac12+frac13+frac14+cdots\ $$
                        $$
                        A=underbrace{(frac{1}{1}+frac{1}{2}+frac{1}{3}+cdots+frac{1}{9})}_{color{red} {9- terms}}
                        +underbrace{(frac{1}{10}+frac{1}{11}+frac{1}{12}+cdots+frac{1}{99})}_{color{red} {90- terms}}\+underbrace{(frac{1}{101}+frac{1}{102}+frac{1}{103}+cdots+frac{1}{999})}_{color{red} {900- terms}}+cdots \ to $$
                        $$\A>9 times(frac{1}{10})+(99-10+1)times frac{1}{100}+(999-100+1)times frac{1}{1000}+... \A>frac{9}{10}+frac{90}{100}+frac{90}{100}+frac{900}{1000}+...\ to A>underbrace{frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+...}_{color{red} {text{ m group} ,text{ and} space mto infty}} to infty
                        $$



                        Showing that $A$ diverges by grouping numbers.






                        share|cite|improve this answer

























                          up vote
                          23
                          down vote










                          up vote
                          23
                          down vote









                          Let's group the terms as follows:$$A=frac11+frac12+frac13+frac14+cdots\ $$
                          $$
                          A=underbrace{(frac{1}{1}+frac{1}{2}+frac{1}{3}+cdots+frac{1}{9})}_{color{red} {9- terms}}
                          +underbrace{(frac{1}{10}+frac{1}{11}+frac{1}{12}+cdots+frac{1}{99})}_{color{red} {90- terms}}\+underbrace{(frac{1}{101}+frac{1}{102}+frac{1}{103}+cdots+frac{1}{999})}_{color{red} {900- terms}}+cdots \ to $$
                          $$\A>9 times(frac{1}{10})+(99-10+1)times frac{1}{100}+(999-100+1)times frac{1}{1000}+... \A>frac{9}{10}+frac{90}{100}+frac{90}{100}+frac{900}{1000}+...\ to A>underbrace{frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+...}_{color{red} {text{ m group} ,text{ and} space mto infty}} to infty
                          $$



                          Showing that $A$ diverges by grouping numbers.






                          share|cite|improve this answer














                          Let's group the terms as follows:$$A=frac11+frac12+frac13+frac14+cdots\ $$
                          $$
                          A=underbrace{(frac{1}{1}+frac{1}{2}+frac{1}{3}+cdots+frac{1}{9})}_{color{red} {9- terms}}
                          +underbrace{(frac{1}{10}+frac{1}{11}+frac{1}{12}+cdots+frac{1}{99})}_{color{red} {90- terms}}\+underbrace{(frac{1}{101}+frac{1}{102}+frac{1}{103}+cdots+frac{1}{999})}_{color{red} {900- terms}}+cdots \ to $$
                          $$\A>9 times(frac{1}{10})+(99-10+1)times frac{1}{100}+(999-100+1)times frac{1}{1000}+... \A>frac{9}{10}+frac{90}{100}+frac{90}{100}+frac{900}{1000}+...\ to A>underbrace{frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+frac{9}{10}+...}_{color{red} {text{ m group} ,text{ and} space mto infty}} to infty
                          $$



                          Showing that $A$ diverges by grouping numbers.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jul 28 '17 at 8:31

























                          answered Nov 1 '13 at 17:07









                          Khosrotash

                          16.8k12361




                          16.8k12361






















                              up vote
                              22
                              down vote













                              The answer given by AgCl is a classic one. And possibly pedagogically best; I don't know.



                              I also like the following argument. I'm not sure what students who are new to the topic will think about it.



                              Suppose 1 + 1/2 + 1/3 + 1/4 + ... adds up to some finite total S. Now group terms in the following way:



                              $$1 + frac{1}{2} > frac{1}{2} + frac{1}{2} = frac{2}{2} = 1$$



                              $$frac{1}{3} + frac{1}{4} > frac{1}{4} + frac{1}{4} = frac{2}{4} = frac{1}{2}$$



                              $$frac{1}{5} + frac{1}{6} > frac{1}{6} + frac{1}{6} = frac{2}{6} = frac{1}{3}$$



                              Continuing in this way, we get $S > S$, a contradiction.






                              share|cite|improve this answer



















                              • 1




                                Not really. From $S_n > T_n$ you can only conclude that $lim S_n ge lim T_n$.
                                – lhf
                                Jul 10 '11 at 21:24








                              • 6




                                @lhf: That's right, but that can be easily fixed here (with $S_n = 1 + 1/2 + dots + 1/2n$ and $T_n = 1 + 1/2 + dots + 1/n$): we can use a better inequality, like say $S_n ge T_n + 1/2$ (using just the first step) to conclude that $lim S_n ge lim T_n + 1/2$, contradicting $S = lim S_n = lim T_n$.
                                – ShreevatsaR
                                Jul 11 '11 at 4:18















                              up vote
                              22
                              down vote













                              The answer given by AgCl is a classic one. And possibly pedagogically best; I don't know.



                              I also like the following argument. I'm not sure what students who are new to the topic will think about it.



                              Suppose 1 + 1/2 + 1/3 + 1/4 + ... adds up to some finite total S. Now group terms in the following way:



                              $$1 + frac{1}{2} > frac{1}{2} + frac{1}{2} = frac{2}{2} = 1$$



                              $$frac{1}{3} + frac{1}{4} > frac{1}{4} + frac{1}{4} = frac{2}{4} = frac{1}{2}$$



                              $$frac{1}{5} + frac{1}{6} > frac{1}{6} + frac{1}{6} = frac{2}{6} = frac{1}{3}$$



                              Continuing in this way, we get $S > S$, a contradiction.






                              share|cite|improve this answer



















                              • 1




                                Not really. From $S_n > T_n$ you can only conclude that $lim S_n ge lim T_n$.
                                – lhf
                                Jul 10 '11 at 21:24








                              • 6




                                @lhf: That's right, but that can be easily fixed here (with $S_n = 1 + 1/2 + dots + 1/2n$ and $T_n = 1 + 1/2 + dots + 1/n$): we can use a better inequality, like say $S_n ge T_n + 1/2$ (using just the first step) to conclude that $lim S_n ge lim T_n + 1/2$, contradicting $S = lim S_n = lim T_n$.
                                – ShreevatsaR
                                Jul 11 '11 at 4:18













                              up vote
                              22
                              down vote










                              up vote
                              22
                              down vote









                              The answer given by AgCl is a classic one. And possibly pedagogically best; I don't know.



                              I also like the following argument. I'm not sure what students who are new to the topic will think about it.



                              Suppose 1 + 1/2 + 1/3 + 1/4 + ... adds up to some finite total S. Now group terms in the following way:



                              $$1 + frac{1}{2} > frac{1}{2} + frac{1}{2} = frac{2}{2} = 1$$



                              $$frac{1}{3} + frac{1}{4} > frac{1}{4} + frac{1}{4} = frac{2}{4} = frac{1}{2}$$



                              $$frac{1}{5} + frac{1}{6} > frac{1}{6} + frac{1}{6} = frac{2}{6} = frac{1}{3}$$



                              Continuing in this way, we get $S > S$, a contradiction.






                              share|cite|improve this answer














                              The answer given by AgCl is a classic one. And possibly pedagogically best; I don't know.



                              I also like the following argument. I'm not sure what students who are new to the topic will think about it.



                              Suppose 1 + 1/2 + 1/3 + 1/4 + ... adds up to some finite total S. Now group terms in the following way:



                              $$1 + frac{1}{2} > frac{1}{2} + frac{1}{2} = frac{2}{2} = 1$$



                              $$frac{1}{3} + frac{1}{4} > frac{1}{4} + frac{1}{4} = frac{2}{4} = frac{1}{2}$$



                              $$frac{1}{5} + frac{1}{6} > frac{1}{6} + frac{1}{6} = frac{2}{6} = frac{1}{3}$$



                              Continuing in this way, we get $S > S$, a contradiction.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Mar 15 '13 at 20:15









                              Dominic Michaelis

                              17.6k43570




                              17.6k43570










                              answered Jul 10 '11 at 21:16









                              idmercer

                              1,3231225




                              1,3231225








                              • 1




                                Not really. From $S_n > T_n$ you can only conclude that $lim S_n ge lim T_n$.
                                – lhf
                                Jul 10 '11 at 21:24








                              • 6




                                @lhf: That's right, but that can be easily fixed here (with $S_n = 1 + 1/2 + dots + 1/2n$ and $T_n = 1 + 1/2 + dots + 1/n$): we can use a better inequality, like say $S_n ge T_n + 1/2$ (using just the first step) to conclude that $lim S_n ge lim T_n + 1/2$, contradicting $S = lim S_n = lim T_n$.
                                – ShreevatsaR
                                Jul 11 '11 at 4:18














                              • 1




                                Not really. From $S_n > T_n$ you can only conclude that $lim S_n ge lim T_n$.
                                – lhf
                                Jul 10 '11 at 21:24








                              • 6




                                @lhf: That's right, but that can be easily fixed here (with $S_n = 1 + 1/2 + dots + 1/2n$ and $T_n = 1 + 1/2 + dots + 1/n$): we can use a better inequality, like say $S_n ge T_n + 1/2$ (using just the first step) to conclude that $lim S_n ge lim T_n + 1/2$, contradicting $S = lim S_n = lim T_n$.
                                – ShreevatsaR
                                Jul 11 '11 at 4:18








                              1




                              1




                              Not really. From $S_n > T_n$ you can only conclude that $lim S_n ge lim T_n$.
                              – lhf
                              Jul 10 '11 at 21:24






                              Not really. From $S_n > T_n$ you can only conclude that $lim S_n ge lim T_n$.
                              – lhf
                              Jul 10 '11 at 21:24






                              6




                              6




                              @lhf: That's right, but that can be easily fixed here (with $S_n = 1 + 1/2 + dots + 1/2n$ and $T_n = 1 + 1/2 + dots + 1/n$): we can use a better inequality, like say $S_n ge T_n + 1/2$ (using just the first step) to conclude that $lim S_n ge lim T_n + 1/2$, contradicting $S = lim S_n = lim T_n$.
                              – ShreevatsaR
                              Jul 11 '11 at 4:18




                              @lhf: That's right, but that can be easily fixed here (with $S_n = 1 + 1/2 + dots + 1/2n$ and $T_n = 1 + 1/2 + dots + 1/n$): we can use a better inequality, like say $S_n ge T_n + 1/2$ (using just the first step) to conclude that $lim S_n ge lim T_n + 1/2$, contradicting $S = lim S_n = lim T_n$.
                              – ShreevatsaR
                              Jul 11 '11 at 4:18










                              up vote
                              19
                              down vote













                              An alternative proof (translated and adapted from this comment by Filipe Oliveira, in Portuguese, posted also here). Let $ f(x)=ln(1+x)$. Then $f'(x)=dfrac {1}{1+x}$ and $ f'(0)=1$. Hence



                              $$displaystylelim_{xto 0}dfrac{ln(1+x)}{x}=lim_{xto 0}dfrac{ln(1+x)-ln(1)}{x-0}=1,$$



                              and



                              $$ displaystylelim_{ntoinfty} dfrac{lnleft(1+dfrac{1}{n}right)}{dfrac {1}{n}}=1>0.$$



                              So, the series $displaystylesumdfrac{1}{n}$ and $displaystylesumlnleft(1+dfrac {1}{n}right)$ are both convergent or divergent. Since



                              $$lnleft(1+dfrac {1}{n}right)=lnleft(dfrac{n+1}{n}right)=ln (n+1)-ln(n),$$



                              we have



                              $$displaystylesum_{n=1}^Nlnleft(1+dfrac {1}{n}right)=ln(N+1)-ln(1)=ln(N+1).$$



                              Thus $displaystylesum_{n=1}^{infty}lnleft(1+dfrac {1}{n}right)$ is divergent and so is $displaystylesum_{n=1}^{infty}dfrac{1}{n}$.






                              share|cite|improve this answer

























                                up vote
                                19
                                down vote













                                An alternative proof (translated and adapted from this comment by Filipe Oliveira, in Portuguese, posted also here). Let $ f(x)=ln(1+x)$. Then $f'(x)=dfrac {1}{1+x}$ and $ f'(0)=1$. Hence



                                $$displaystylelim_{xto 0}dfrac{ln(1+x)}{x}=lim_{xto 0}dfrac{ln(1+x)-ln(1)}{x-0}=1,$$



                                and



                                $$ displaystylelim_{ntoinfty} dfrac{lnleft(1+dfrac{1}{n}right)}{dfrac {1}{n}}=1>0.$$



                                So, the series $displaystylesumdfrac{1}{n}$ and $displaystylesumlnleft(1+dfrac {1}{n}right)$ are both convergent or divergent. Since



                                $$lnleft(1+dfrac {1}{n}right)=lnleft(dfrac{n+1}{n}right)=ln (n+1)-ln(n),$$



                                we have



                                $$displaystylesum_{n=1}^Nlnleft(1+dfrac {1}{n}right)=ln(N+1)-ln(1)=ln(N+1).$$



                                Thus $displaystylesum_{n=1}^{infty}lnleft(1+dfrac {1}{n}right)$ is divergent and so is $displaystylesum_{n=1}^{infty}dfrac{1}{n}$.






                                share|cite|improve this answer























                                  up vote
                                  19
                                  down vote










                                  up vote
                                  19
                                  down vote









                                  An alternative proof (translated and adapted from this comment by Filipe Oliveira, in Portuguese, posted also here). Let $ f(x)=ln(1+x)$. Then $f'(x)=dfrac {1}{1+x}$ and $ f'(0)=1$. Hence



                                  $$displaystylelim_{xto 0}dfrac{ln(1+x)}{x}=lim_{xto 0}dfrac{ln(1+x)-ln(1)}{x-0}=1,$$



                                  and



                                  $$ displaystylelim_{ntoinfty} dfrac{lnleft(1+dfrac{1}{n}right)}{dfrac {1}{n}}=1>0.$$



                                  So, the series $displaystylesumdfrac{1}{n}$ and $displaystylesumlnleft(1+dfrac {1}{n}right)$ are both convergent or divergent. Since



                                  $$lnleft(1+dfrac {1}{n}right)=lnleft(dfrac{n+1}{n}right)=ln (n+1)-ln(n),$$



                                  we have



                                  $$displaystylesum_{n=1}^Nlnleft(1+dfrac {1}{n}right)=ln(N+1)-ln(1)=ln(N+1).$$



                                  Thus $displaystylesum_{n=1}^{infty}lnleft(1+dfrac {1}{n}right)$ is divergent and so is $displaystylesum_{n=1}^{infty}dfrac{1}{n}$.






                                  share|cite|improve this answer












                                  An alternative proof (translated and adapted from this comment by Filipe Oliveira, in Portuguese, posted also here). Let $ f(x)=ln(1+x)$. Then $f'(x)=dfrac {1}{1+x}$ and $ f'(0)=1$. Hence



                                  $$displaystylelim_{xto 0}dfrac{ln(1+x)}{x}=lim_{xto 0}dfrac{ln(1+x)-ln(1)}{x-0}=1,$$



                                  and



                                  $$ displaystylelim_{ntoinfty} dfrac{lnleft(1+dfrac{1}{n}right)}{dfrac {1}{n}}=1>0.$$



                                  So, the series $displaystylesumdfrac{1}{n}$ and $displaystylesumlnleft(1+dfrac {1}{n}right)$ are both convergent or divergent. Since



                                  $$lnleft(1+dfrac {1}{n}right)=lnleft(dfrac{n+1}{n}right)=ln (n+1)-ln(n),$$



                                  we have



                                  $$displaystylesum_{n=1}^Nlnleft(1+dfrac {1}{n}right)=ln(N+1)-ln(1)=ln(N+1).$$



                                  Thus $displaystylesum_{n=1}^{infty}lnleft(1+dfrac {1}{n}right)$ is divergent and so is $displaystylesum_{n=1}^{infty}dfrac{1}{n}$.







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Jul 10 '11 at 21:50









                                  Américo Tavares

                                  32.2k1079202




                                  32.2k1079202






















                                      up vote
                                      19
                                      down vote













                                      This is not as good an answer as AgCl's, nonetheless people may find it interesting.



                                      If you're used to calculus then you might notice that the sum $$
                                      1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n}$$ is very close to the integral from $1$ to $n$ of $frac{1}{x}$. This definite integral is ln(n), so you should expect $1+frac{1}{2}+frac{1}{3}+dots+
                                      frac{1}{n}$ to grow like $ln(n)$.



                                      Although this argument can be made rigorous, it's still unsatisfying because it depends on the fact that the derivative of $ln(x)$ is $frac{1}{x}$, which is probably harder than the original question. Nonetheless it does illustrate a good general heuristic for quickly determining how sums behave if you already know calculus.






                                      share|cite|improve this answer



















                                      • 2




                                        If you look at a Riemann sum for intervals with width 1, you can pretty quickly see that the integral of 1/x from 1 to infinity must be less than the sum of the harmonic series.
                                        – Isaac
                                        Jul 21 '10 at 5:51










                                      • Thank you for adding this answer. I was hoping to avoid an answer that involved integration, so I also prefer AgCl's answer. But I am happy to see more than one demonstration/proof.
                                        – bryn
                                        Jul 22 '10 at 11:33










                                      • The sum is closer to the integral from $frac{1}{2}$ to $n+frac{1}{2}$ of $frac{1}{x}$, which is $log(2n+1)$ math.stackexchange.com/a/1602945/134791
                                        – Jaume Oliver Lafont
                                        Jan 25 '16 at 23:02















                                      up vote
                                      19
                                      down vote













                                      This is not as good an answer as AgCl's, nonetheless people may find it interesting.



                                      If you're used to calculus then you might notice that the sum $$
                                      1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n}$$ is very close to the integral from $1$ to $n$ of $frac{1}{x}$. This definite integral is ln(n), so you should expect $1+frac{1}{2}+frac{1}{3}+dots+
                                      frac{1}{n}$ to grow like $ln(n)$.



                                      Although this argument can be made rigorous, it's still unsatisfying because it depends on the fact that the derivative of $ln(x)$ is $frac{1}{x}$, which is probably harder than the original question. Nonetheless it does illustrate a good general heuristic for quickly determining how sums behave if you already know calculus.






                                      share|cite|improve this answer



















                                      • 2




                                        If you look at a Riemann sum for intervals with width 1, you can pretty quickly see that the integral of 1/x from 1 to infinity must be less than the sum of the harmonic series.
                                        – Isaac
                                        Jul 21 '10 at 5:51










                                      • Thank you for adding this answer. I was hoping to avoid an answer that involved integration, so I also prefer AgCl's answer. But I am happy to see more than one demonstration/proof.
                                        – bryn
                                        Jul 22 '10 at 11:33










                                      • The sum is closer to the integral from $frac{1}{2}$ to $n+frac{1}{2}$ of $frac{1}{x}$, which is $log(2n+1)$ math.stackexchange.com/a/1602945/134791
                                        – Jaume Oliver Lafont
                                        Jan 25 '16 at 23:02













                                      up vote
                                      19
                                      down vote










                                      up vote
                                      19
                                      down vote









                                      This is not as good an answer as AgCl's, nonetheless people may find it interesting.



                                      If you're used to calculus then you might notice that the sum $$
                                      1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n}$$ is very close to the integral from $1$ to $n$ of $frac{1}{x}$. This definite integral is ln(n), so you should expect $1+frac{1}{2}+frac{1}{3}+dots+
                                      frac{1}{n}$ to grow like $ln(n)$.



                                      Although this argument can be made rigorous, it's still unsatisfying because it depends on the fact that the derivative of $ln(x)$ is $frac{1}{x}$, which is probably harder than the original question. Nonetheless it does illustrate a good general heuristic for quickly determining how sums behave if you already know calculus.






                                      share|cite|improve this answer














                                      This is not as good an answer as AgCl's, nonetheless people may find it interesting.



                                      If you're used to calculus then you might notice that the sum $$
                                      1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n}$$ is very close to the integral from $1$ to $n$ of $frac{1}{x}$. This definite integral is ln(n), so you should expect $1+frac{1}{2}+frac{1}{3}+dots+
                                      frac{1}{n}$ to grow like $ln(n)$.



                                      Although this argument can be made rigorous, it's still unsatisfying because it depends on the fact that the derivative of $ln(x)$ is $frac{1}{x}$, which is probably harder than the original question. Nonetheless it does illustrate a good general heuristic for quickly determining how sums behave if you already know calculus.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Mar 15 '13 at 19:33









                                      Dominic Michaelis

                                      17.6k43570




                                      17.6k43570










                                      answered Jul 21 '10 at 5:28









                                      Noah Snyder

                                      7,47722854




                                      7,47722854








                                      • 2




                                        If you look at a Riemann sum for intervals with width 1, you can pretty quickly see that the integral of 1/x from 1 to infinity must be less than the sum of the harmonic series.
                                        – Isaac
                                        Jul 21 '10 at 5:51










                                      • Thank you for adding this answer. I was hoping to avoid an answer that involved integration, so I also prefer AgCl's answer. But I am happy to see more than one demonstration/proof.
                                        – bryn
                                        Jul 22 '10 at 11:33










                                      • The sum is closer to the integral from $frac{1}{2}$ to $n+frac{1}{2}$ of $frac{1}{x}$, which is $log(2n+1)$ math.stackexchange.com/a/1602945/134791
                                        – Jaume Oliver Lafont
                                        Jan 25 '16 at 23:02














                                      • 2




                                        If you look at a Riemann sum for intervals with width 1, you can pretty quickly see that the integral of 1/x from 1 to infinity must be less than the sum of the harmonic series.
                                        – Isaac
                                        Jul 21 '10 at 5:51










                                      • Thank you for adding this answer. I was hoping to avoid an answer that involved integration, so I also prefer AgCl's answer. But I am happy to see more than one demonstration/proof.
                                        – bryn
                                        Jul 22 '10 at 11:33










                                      • The sum is closer to the integral from $frac{1}{2}$ to $n+frac{1}{2}$ of $frac{1}{x}$, which is $log(2n+1)$ math.stackexchange.com/a/1602945/134791
                                        – Jaume Oliver Lafont
                                        Jan 25 '16 at 23:02








                                      2




                                      2




                                      If you look at a Riemann sum for intervals with width 1, you can pretty quickly see that the integral of 1/x from 1 to infinity must be less than the sum of the harmonic series.
                                      – Isaac
                                      Jul 21 '10 at 5:51




                                      If you look at a Riemann sum for intervals with width 1, you can pretty quickly see that the integral of 1/x from 1 to infinity must be less than the sum of the harmonic series.
                                      – Isaac
                                      Jul 21 '10 at 5:51












                                      Thank you for adding this answer. I was hoping to avoid an answer that involved integration, so I also prefer AgCl's answer. But I am happy to see more than one demonstration/proof.
                                      – bryn
                                      Jul 22 '10 at 11:33




                                      Thank you for adding this answer. I was hoping to avoid an answer that involved integration, so I also prefer AgCl's answer. But I am happy to see more than one demonstration/proof.
                                      – bryn
                                      Jul 22 '10 at 11:33












                                      The sum is closer to the integral from $frac{1}{2}$ to $n+frac{1}{2}$ of $frac{1}{x}$, which is $log(2n+1)$ math.stackexchange.com/a/1602945/134791
                                      – Jaume Oliver Lafont
                                      Jan 25 '16 at 23:02




                                      The sum is closer to the integral from $frac{1}{2}$ to $n+frac{1}{2}$ of $frac{1}{x}$, which is $log(2n+1)$ math.stackexchange.com/a/1602945/134791
                                      – Jaume Oliver Lafont
                                      Jan 25 '16 at 23:02










                                      up vote
                                      17
                                      down vote













                                      This was bumped, so I'll add a proof sweet proof I saw in this site. Exponentiate $H_n$ and get $$e^{H_n}=prod_{k=1}^n e^{1/k}gtprod_{k=1}^nleft(1+frac{1}{k}right)=n+1.$$ Therefore, $H_ngtlog(n+1)$, so we are done. We used $e^xgt1+x$ and telescoped the resulting product.






                                      share|cite|improve this answer





















                                      • Oh, that's unique.
                                        – Simply Beautiful Art
                                        Oct 7 '16 at 1:12















                                      up vote
                                      17
                                      down vote













                                      This was bumped, so I'll add a proof sweet proof I saw in this site. Exponentiate $H_n$ and get $$e^{H_n}=prod_{k=1}^n e^{1/k}gtprod_{k=1}^nleft(1+frac{1}{k}right)=n+1.$$ Therefore, $H_ngtlog(n+1)$, so we are done. We used $e^xgt1+x$ and telescoped the resulting product.






                                      share|cite|improve this answer





















                                      • Oh, that's unique.
                                        – Simply Beautiful Art
                                        Oct 7 '16 at 1:12













                                      up vote
                                      17
                                      down vote










                                      up vote
                                      17
                                      down vote









                                      This was bumped, so I'll add a proof sweet proof I saw in this site. Exponentiate $H_n$ and get $$e^{H_n}=prod_{k=1}^n e^{1/k}gtprod_{k=1}^nleft(1+frac{1}{k}right)=n+1.$$ Therefore, $H_ngtlog(n+1)$, so we are done. We used $e^xgt1+x$ and telescoped the resulting product.






                                      share|cite|improve this answer












                                      This was bumped, so I'll add a proof sweet proof I saw in this site. Exponentiate $H_n$ and get $$e^{H_n}=prod_{k=1}^n e^{1/k}gtprod_{k=1}^nleft(1+frac{1}{k}right)=n+1.$$ Therefore, $H_ngtlog(n+1)$, so we are done. We used $e^xgt1+x$ and telescoped the resulting product.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 27 '13 at 14:49









                                      Ian Mateus

                                      4,66032452




                                      4,66032452












                                      • Oh, that's unique.
                                        – Simply Beautiful Art
                                        Oct 7 '16 at 1:12


















                                      • Oh, that's unique.
                                        – Simply Beautiful Art
                                        Oct 7 '16 at 1:12
















                                      Oh, that's unique.
                                      – Simply Beautiful Art
                                      Oct 7 '16 at 1:12




                                      Oh, that's unique.
                                      – Simply Beautiful Art
                                      Oct 7 '16 at 1:12










                                      up vote
                                      13
                                      down vote













                                      Another interesting proof is based upon one of the consequences of the Lagrange's theorem applied on $ln(x)$ function, namely:



                                      $$frac{1}{k+1} < ln(k+1)-ln(k)<frac{1}{k} space , space kinmathbb{N} ,space k>0$$



                                      Taking $k=1,2,...,n$ values to the inequality and then summing all relations, we get the required result.



                                      The proof is complete.






                                      share|cite|improve this answer



























                                        up vote
                                        13
                                        down vote













                                        Another interesting proof is based upon one of the consequences of the Lagrange's theorem applied on $ln(x)$ function, namely:



                                        $$frac{1}{k+1} < ln(k+1)-ln(k)<frac{1}{k} space , space kinmathbb{N} ,space k>0$$



                                        Taking $k=1,2,...,n$ values to the inequality and then summing all relations, we get the required result.



                                        The proof is complete.






                                        share|cite|improve this answer

























                                          up vote
                                          13
                                          down vote










                                          up vote
                                          13
                                          down vote









                                          Another interesting proof is based upon one of the consequences of the Lagrange's theorem applied on $ln(x)$ function, namely:



                                          $$frac{1}{k+1} < ln(k+1)-ln(k)<frac{1}{k} space , space kinmathbb{N} ,space k>0$$



                                          Taking $k=1,2,...,n$ values to the inequality and then summing all relations, we get the required result.



                                          The proof is complete.






                                          share|cite|improve this answer














                                          Another interesting proof is based upon one of the consequences of the Lagrange's theorem applied on $ln(x)$ function, namely:



                                          $$frac{1}{k+1} < ln(k+1)-ln(k)<frac{1}{k} space , space kinmathbb{N} ,space k>0$$



                                          Taking $k=1,2,...,n$ values to the inequality and then summing all relations, we get the required result.



                                          The proof is complete.







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          edited Oct 26 '12 at 13:03

























                                          answered May 29 '12 at 13:52









                                          user 1357113

                                          22.2k875224




                                          22.2k875224






















                                              up vote
                                              11
                                              down vote













                                              There also exists a proof for the divergence of the harmonic series that involves the Integral Test. It goes as follows.



                                              It is possible to prove that the harmonic series diverges by comparing its sum with an improper integral. Specifically, consider the arrangement of rectangles shown in the figure of $ y = dfrac {1}{x} $:



                                              y=1/x



                                              Each rectangle is $1$ unit wide and $frac{1}{n}$ units high, so the total area of the rectangles is the sum of the harmonic series: $$ displaystylesum left( text {enclosed rectangle are} right) = displaystylesum_{k=1}^{infty} dfrac {1}{k}. $$Now, the total area under the curve is given by $$ displaystyleint_{1}^{infty} dfrac {mathrm{d}x}{x} = infty. $$Since this area is entirely contained within the rectangles, the total area of the rectangles must be infinite as well. More precisely, this proves that $$ displaystylesum_{n=1}^{k} dfrac {1}{n} > displaystyleint_{1}^{k+1} dfrac {mathrm{d}x}{x} = ln (k+1). $$This is the backbone of what we know today as the integral test.



                                              Interestingly, the alternating harmonic series does converge: $$ displaystylesum_{n=1}^{infty} dfrac {(-1)^n}{n} = ln 2. $$And so does the $p$-harmonic series with $p>1$.






                                              share|cite|improve this answer



























                                                up vote
                                                11
                                                down vote













                                                There also exists a proof for the divergence of the harmonic series that involves the Integral Test. It goes as follows.



                                                It is possible to prove that the harmonic series diverges by comparing its sum with an improper integral. Specifically, consider the arrangement of rectangles shown in the figure of $ y = dfrac {1}{x} $:



                                                y=1/x



                                                Each rectangle is $1$ unit wide and $frac{1}{n}$ units high, so the total area of the rectangles is the sum of the harmonic series: $$ displaystylesum left( text {enclosed rectangle are} right) = displaystylesum_{k=1}^{infty} dfrac {1}{k}. $$Now, the total area under the curve is given by $$ displaystyleint_{1}^{infty} dfrac {mathrm{d}x}{x} = infty. $$Since this area is entirely contained within the rectangles, the total area of the rectangles must be infinite as well. More precisely, this proves that $$ displaystylesum_{n=1}^{k} dfrac {1}{n} > displaystyleint_{1}^{k+1} dfrac {mathrm{d}x}{x} = ln (k+1). $$This is the backbone of what we know today as the integral test.



                                                Interestingly, the alternating harmonic series does converge: $$ displaystylesum_{n=1}^{infty} dfrac {(-1)^n}{n} = ln 2. $$And so does the $p$-harmonic series with $p>1$.






                                                share|cite|improve this answer

























                                                  up vote
                                                  11
                                                  down vote










                                                  up vote
                                                  11
                                                  down vote









                                                  There also exists a proof for the divergence of the harmonic series that involves the Integral Test. It goes as follows.



                                                  It is possible to prove that the harmonic series diverges by comparing its sum with an improper integral. Specifically, consider the arrangement of rectangles shown in the figure of $ y = dfrac {1}{x} $:



                                                  y=1/x



                                                  Each rectangle is $1$ unit wide and $frac{1}{n}$ units high, so the total area of the rectangles is the sum of the harmonic series: $$ displaystylesum left( text {enclosed rectangle are} right) = displaystylesum_{k=1}^{infty} dfrac {1}{k}. $$Now, the total area under the curve is given by $$ displaystyleint_{1}^{infty} dfrac {mathrm{d}x}{x} = infty. $$Since this area is entirely contained within the rectangles, the total area of the rectangles must be infinite as well. More precisely, this proves that $$ displaystylesum_{n=1}^{k} dfrac {1}{n} > displaystyleint_{1}^{k+1} dfrac {mathrm{d}x}{x} = ln (k+1). $$This is the backbone of what we know today as the integral test.



                                                  Interestingly, the alternating harmonic series does converge: $$ displaystylesum_{n=1}^{infty} dfrac {(-1)^n}{n} = ln 2. $$And so does the $p$-harmonic series with $p>1$.






                                                  share|cite|improve this answer














                                                  There also exists a proof for the divergence of the harmonic series that involves the Integral Test. It goes as follows.



                                                  It is possible to prove that the harmonic series diverges by comparing its sum with an improper integral. Specifically, consider the arrangement of rectangles shown in the figure of $ y = dfrac {1}{x} $:



                                                  y=1/x



                                                  Each rectangle is $1$ unit wide and $frac{1}{n}$ units high, so the total area of the rectangles is the sum of the harmonic series: $$ displaystylesum left( text {enclosed rectangle are} right) = displaystylesum_{k=1}^{infty} dfrac {1}{k}. $$Now, the total area under the curve is given by $$ displaystyleint_{1}^{infty} dfrac {mathrm{d}x}{x} = infty. $$Since this area is entirely contained within the rectangles, the total area of the rectangles must be infinite as well. More precisely, this proves that $$ displaystylesum_{n=1}^{k} dfrac {1}{n} > displaystyleint_{1}^{k+1} dfrac {mathrm{d}x}{x} = ln (k+1). $$This is the backbone of what we know today as the integral test.



                                                  Interestingly, the alternating harmonic series does converge: $$ displaystylesum_{n=1}^{infty} dfrac {(-1)^n}{n} = ln 2. $$And so does the $p$-harmonic series with $p>1$.







                                                  share|cite|improve this answer














                                                  share|cite|improve this answer



                                                  share|cite|improve this answer








                                                  edited Feb 8 '15 at 21:45









                                                  Cyclohexanol.

                                                  7,06911753




                                                  7,06911753










                                                  answered Nov 1 '13 at 17:38









                                                  Ahaan S. Rungta

                                                  6,46052160




                                                  6,46052160






















                                                      up vote
                                                      8
                                                      down vote













                                                      Let's assume that $sum_{n=1}^{infty}frac1n=:Hin mathbb{R}$, then
                                                      $$H=frac11+frac12+frac13+frac14+frac15+frac16 +ldots $$
                                                      $$Hgeqslant frac11+frac12 +frac14+frac14+frac16+frac16+ldots$$
                                                      $$Hgeqslant frac11+frac12+frac12+frac13+frac14+frac15+ldots$$
                                                      $$Hgeqslant frac12 +H Rightarrow 0geqslant frac12$$
                                                      Since the last inequality doesn't hold, we can conclude that the sum doesn't converge.






                                                      share|cite|improve this answer

























                                                        up vote
                                                        8
                                                        down vote













                                                        Let's assume that $sum_{n=1}^{infty}frac1n=:Hin mathbb{R}$, then
                                                        $$H=frac11+frac12+frac13+frac14+frac15+frac16 +ldots $$
                                                        $$Hgeqslant frac11+frac12 +frac14+frac14+frac16+frac16+ldots$$
                                                        $$Hgeqslant frac11+frac12+frac12+frac13+frac14+frac15+ldots$$
                                                        $$Hgeqslant frac12 +H Rightarrow 0geqslant frac12$$
                                                        Since the last inequality doesn't hold, we can conclude that the sum doesn't converge.






                                                        share|cite|improve this answer























                                                          up vote
                                                          8
                                                          down vote










                                                          up vote
                                                          8
                                                          down vote









                                                          Let's assume that $sum_{n=1}^{infty}frac1n=:Hin mathbb{R}$, then
                                                          $$H=frac11+frac12+frac13+frac14+frac15+frac16 +ldots $$
                                                          $$Hgeqslant frac11+frac12 +frac14+frac14+frac16+frac16+ldots$$
                                                          $$Hgeqslant frac11+frac12+frac12+frac13+frac14+frac15+ldots$$
                                                          $$Hgeqslant frac12 +H Rightarrow 0geqslant frac12$$
                                                          Since the last inequality doesn't hold, we can conclude that the sum doesn't converge.






                                                          share|cite|improve this answer












                                                          Let's assume that $sum_{n=1}^{infty}frac1n=:Hin mathbb{R}$, then
                                                          $$H=frac11+frac12+frac13+frac14+frac15+frac16 +ldots $$
                                                          $$Hgeqslant frac11+frac12 +frac14+frac14+frac16+frac16+ldots$$
                                                          $$Hgeqslant frac11+frac12+frac12+frac13+frac14+frac15+ldots$$
                                                          $$Hgeqslant frac12 +H Rightarrow 0geqslant frac12$$
                                                          Since the last inequality doesn't hold, we can conclude that the sum doesn't converge.







                                                          share|cite|improve this answer












                                                          share|cite|improve this answer



                                                          share|cite|improve this answer










                                                          answered Feb 8 '15 at 22:01









                                                          Rasmus Erlemann

                                                          2,32421325




                                                          2,32421325






















                                                              up vote
                                                              8
                                                              down vote













                                                              $$int_{0}^{infty}e^{-nx}dx=frac1n$$



                                                              $$sum_{n=1}^{infty}int_{0}^{infty}e^{-nx}dx=lim_{ m to infty}sum_{n=1}^{m}frac1n$$



                                                              using the law of Geometric series



                                                              $$int_{0}^{infty}(frac{1}{1-e^{-x}}-1)dx=lim_{ m to infty}H_m$$



                                                              $$lim_{ m to infty}H_m=left [ ln(e^x-1)-x right ]_0^{infty}toinfty$$






                                                              share|cite|improve this answer























                                                              • Hm, the lower bound goes to $-infty$ it appears.
                                                                – Simply Beautiful Art
                                                                Oct 6 '16 at 16:40










                                                              • @SimpleArt The upper bound goes to 0 and The Lower goes to $+infty$ ,,,$-ln 0^+$
                                                                – mhd.math
                                                                Oct 7 '16 at 11:20












                                                              • Oh right, duh, didn't quite use that FTOC correctly.
                                                                – Simply Beautiful Art
                                                                Oct 7 '16 at 13:19















                                                              up vote
                                                              8
                                                              down vote













                                                              $$int_{0}^{infty}e^{-nx}dx=frac1n$$



                                                              $$sum_{n=1}^{infty}int_{0}^{infty}e^{-nx}dx=lim_{ m to infty}sum_{n=1}^{m}frac1n$$



                                                              using the law of Geometric series



                                                              $$int_{0}^{infty}(frac{1}{1-e^{-x}}-1)dx=lim_{ m to infty}H_m$$



                                                              $$lim_{ m to infty}H_m=left [ ln(e^x-1)-x right ]_0^{infty}toinfty$$






                                                              share|cite|improve this answer























                                                              • Hm, the lower bound goes to $-infty$ it appears.
                                                                – Simply Beautiful Art
                                                                Oct 6 '16 at 16:40










                                                              • @SimpleArt The upper bound goes to 0 and The Lower goes to $+infty$ ,,,$-ln 0^+$
                                                                – mhd.math
                                                                Oct 7 '16 at 11:20












                                                              • Oh right, duh, didn't quite use that FTOC correctly.
                                                                – Simply Beautiful Art
                                                                Oct 7 '16 at 13:19













                                                              up vote
                                                              8
                                                              down vote










                                                              up vote
                                                              8
                                                              down vote









                                                              $$int_{0}^{infty}e^{-nx}dx=frac1n$$



                                                              $$sum_{n=1}^{infty}int_{0}^{infty}e^{-nx}dx=lim_{ m to infty}sum_{n=1}^{m}frac1n$$



                                                              using the law of Geometric series



                                                              $$int_{0}^{infty}(frac{1}{1-e^{-x}}-1)dx=lim_{ m to infty}H_m$$



                                                              $$lim_{ m to infty}H_m=left [ ln(e^x-1)-x right ]_0^{infty}toinfty$$






                                                              share|cite|improve this answer














                                                              $$int_{0}^{infty}e^{-nx}dx=frac1n$$



                                                              $$sum_{n=1}^{infty}int_{0}^{infty}e^{-nx}dx=lim_{ m to infty}sum_{n=1}^{m}frac1n$$



                                                              using the law of Geometric series



                                                              $$int_{0}^{infty}(frac{1}{1-e^{-x}}-1)dx=lim_{ m to infty}H_m$$



                                                              $$lim_{ m to infty}H_m=left [ ln(e^x-1)-x right ]_0^{infty}toinfty$$







                                                              share|cite|improve this answer














                                                              share|cite|improve this answer



                                                              share|cite|improve this answer








                                                              edited Oct 7 '16 at 11:22

























                                                              answered Dec 27 '13 at 14:10









                                                              mhd.math

                                                              3,41711852




                                                              3,41711852












                                                              • Hm, the lower bound goes to $-infty$ it appears.
                                                                – Simply Beautiful Art
                                                                Oct 6 '16 at 16:40










                                                              • @SimpleArt The upper bound goes to 0 and The Lower goes to $+infty$ ,,,$-ln 0^+$
                                                                – mhd.math
                                                                Oct 7 '16 at 11:20












                                                              • Oh right, duh, didn't quite use that FTOC correctly.
                                                                – Simply Beautiful Art
                                                                Oct 7 '16 at 13:19


















                                                              • Hm, the lower bound goes to $-infty$ it appears.
                                                                – Simply Beautiful Art
                                                                Oct 6 '16 at 16:40










                                                              • @SimpleArt The upper bound goes to 0 and The Lower goes to $+infty$ ,,,$-ln 0^+$
                                                                – mhd.math
                                                                Oct 7 '16 at 11:20












                                                              • Oh right, duh, didn't quite use that FTOC correctly.
                                                                – Simply Beautiful Art
                                                                Oct 7 '16 at 13:19
















                                                              Hm, the lower bound goes to $-infty$ it appears.
                                                              – Simply Beautiful Art
                                                              Oct 6 '16 at 16:40




                                                              Hm, the lower bound goes to $-infty$ it appears.
                                                              – Simply Beautiful Art
                                                              Oct 6 '16 at 16:40












                                                              @SimpleArt The upper bound goes to 0 and The Lower goes to $+infty$ ,,,$-ln 0^+$
                                                              – mhd.math
                                                              Oct 7 '16 at 11:20






                                                              @SimpleArt The upper bound goes to 0 and The Lower goes to $+infty$ ,,,$-ln 0^+$
                                                              – mhd.math
                                                              Oct 7 '16 at 11:20














                                                              Oh right, duh, didn't quite use that FTOC correctly.
                                                              – Simply Beautiful Art
                                                              Oct 7 '16 at 13:19




                                                              Oh right, duh, didn't quite use that FTOC correctly.
                                                              – Simply Beautiful Art
                                                              Oct 7 '16 at 13:19










                                                              up vote
                                                              7
                                                              down vote













                                                              Another (different) answer, by the Cauchy Condensation Test :



                                                              $$sum_{n=1}^infty frac{1}{n} < infty iff sum_{n=1}^infty 2^n frac{1}{2^n} = sum_{n=1}^infty 1< infty $$



                                                              The latter is obviously divergent, therefore the former diverges. This is THE shortest proof there is.






                                                              share|cite|improve this answer

























                                                                up vote
                                                                7
                                                                down vote













                                                                Another (different) answer, by the Cauchy Condensation Test :



                                                                $$sum_{n=1}^infty frac{1}{n} < infty iff sum_{n=1}^infty 2^n frac{1}{2^n} = sum_{n=1}^infty 1< infty $$



                                                                The latter is obviously divergent, therefore the former diverges. This is THE shortest proof there is.






                                                                share|cite|improve this answer























                                                                  up vote
                                                                  7
                                                                  down vote










                                                                  up vote
                                                                  7
                                                                  down vote









                                                                  Another (different) answer, by the Cauchy Condensation Test :



                                                                  $$sum_{n=1}^infty frac{1}{n} < infty iff sum_{n=1}^infty 2^n frac{1}{2^n} = sum_{n=1}^infty 1< infty $$



                                                                  The latter is obviously divergent, therefore the former diverges. This is THE shortest proof there is.






                                                                  share|cite|improve this answer












                                                                  Another (different) answer, by the Cauchy Condensation Test :



                                                                  $$sum_{n=1}^infty frac{1}{n} < infty iff sum_{n=1}^infty 2^n frac{1}{2^n} = sum_{n=1}^infty 1< infty $$



                                                                  The latter is obviously divergent, therefore the former diverges. This is THE shortest proof there is.







                                                                  share|cite|improve this answer












                                                                  share|cite|improve this answer



                                                                  share|cite|improve this answer










                                                                  answered Mar 11 '14 at 4:46









                                                                  Squirtle

                                                                  4,1131641




                                                                  4,1131641






















                                                                      up vote
                                                                      7
                                                                      down vote













                                                                      Suppose to the contrary that converges.



                                                                      Let $s_n$ denote the $n$-th partial sum. Since the serie converges, $(s_n)$ is a Cauchy sequence. Let $varepsilon = 1/3$, then there is some $n_0$ such that $|s_q-s_p|< 1/3$ for all $q>pge n_0$. Let $q=2n_0$ and $p=n_0$. Then



                                                                      $$frac{1}{3}>bigg|sum_{n=n_0+1}^{2n_0} frac{1}{n}bigg|gebigg|sum_{n=n_0+1}^{2n_0} frac{1}{2n_0}bigg|=frac{1}{2}$$



                                                                      a contradiction. Then this contradiction shows that the series diverges.






                                                                      share|cite|improve this answer

























                                                                        up vote
                                                                        7
                                                                        down vote













                                                                        Suppose to the contrary that converges.



                                                                        Let $s_n$ denote the $n$-th partial sum. Since the serie converges, $(s_n)$ is a Cauchy sequence. Let $varepsilon = 1/3$, then there is some $n_0$ such that $|s_q-s_p|< 1/3$ for all $q>pge n_0$. Let $q=2n_0$ and $p=n_0$. Then



                                                                        $$frac{1}{3}>bigg|sum_{n=n_0+1}^{2n_0} frac{1}{n}bigg|gebigg|sum_{n=n_0+1}^{2n_0} frac{1}{2n_0}bigg|=frac{1}{2}$$



                                                                        a contradiction. Then this contradiction shows that the series diverges.






                                                                        share|cite|improve this answer























                                                                          up vote
                                                                          7
                                                                          down vote










                                                                          up vote
                                                                          7
                                                                          down vote









                                                                          Suppose to the contrary that converges.



                                                                          Let $s_n$ denote the $n$-th partial sum. Since the serie converges, $(s_n)$ is a Cauchy sequence. Let $varepsilon = 1/3$, then there is some $n_0$ such that $|s_q-s_p|< 1/3$ for all $q>pge n_0$. Let $q=2n_0$ and $p=n_0$. Then



                                                                          $$frac{1}{3}>bigg|sum_{n=n_0+1}^{2n_0} frac{1}{n}bigg|gebigg|sum_{n=n_0+1}^{2n_0} frac{1}{2n_0}bigg|=frac{1}{2}$$



                                                                          a contradiction. Then this contradiction shows that the series diverges.






                                                                          share|cite|improve this answer












                                                                          Suppose to the contrary that converges.



                                                                          Let $s_n$ denote the $n$-th partial sum. Since the serie converges, $(s_n)$ is a Cauchy sequence. Let $varepsilon = 1/3$, then there is some $n_0$ such that $|s_q-s_p|< 1/3$ for all $q>pge n_0$. Let $q=2n_0$ and $p=n_0$. Then



                                                                          $$frac{1}{3}>bigg|sum_{n=n_0+1}^{2n_0} frac{1}{n}bigg|gebigg|sum_{n=n_0+1}^{2n_0} frac{1}{2n_0}bigg|=frac{1}{2}$$



                                                                          a contradiction. Then this contradiction shows that the series diverges.







                                                                          share|cite|improve this answer












                                                                          share|cite|improve this answer



                                                                          share|cite|improve this answer










                                                                          answered Mar 11 '14 at 5:34









                                                                          Jose Antonio

                                                                          4,39421427




                                                                          4,39421427






















                                                                              up vote
                                                                              7
                                                                              down vote













                                                                              enter image description here



                                                                              First suppose $displaystyle A=frac11+frac12+frac13+frac14+cdots$ converges
                                                                              then show that $A>A$. That's paradox.






                                                                              share|cite|improve this answer



















                                                                              • 11




                                                                                Ideally use Latex.
                                                                                – Meow
                                                                                Nov 1 '13 at 17:13








                                                                              • 2




                                                                                I am afraid that this approach is incorrect, since similar versions of it can be applied to convergent infinite series. The reason that it does not work is because the number of terms is infinite. All you've proven is that the second series approaches the final value faster than the first one. But whether this value is ultimately finite or not, you have not shown.
                                                                                – Lucian
                                                                                Jan 14 '15 at 16:32

















                                                                              up vote
                                                                              7
                                                                              down vote













                                                                              enter image description here



                                                                              First suppose $displaystyle A=frac11+frac12+frac13+frac14+cdots$ converges
                                                                              then show that $A>A$. That's paradox.






                                                                              share|cite|improve this answer



















                                                                              • 11




                                                                                Ideally use Latex.
                                                                                – Meow
                                                                                Nov 1 '13 at 17:13








                                                                              • 2




                                                                                I am afraid that this approach is incorrect, since similar versions of it can be applied to convergent infinite series. The reason that it does not work is because the number of terms is infinite. All you've proven is that the second series approaches the final value faster than the first one. But whether this value is ultimately finite or not, you have not shown.
                                                                                – Lucian
                                                                                Jan 14 '15 at 16:32















                                                                              up vote
                                                                              7
                                                                              down vote










                                                                              up vote
                                                                              7
                                                                              down vote









                                                                              enter image description here



                                                                              First suppose $displaystyle A=frac11+frac12+frac13+frac14+cdots$ converges
                                                                              then show that $A>A$. That's paradox.






                                                                              share|cite|improve this answer














                                                                              enter image description here



                                                                              First suppose $displaystyle A=frac11+frac12+frac13+frac14+cdots$ converges
                                                                              then show that $A>A$. That's paradox.







                                                                              share|cite|improve this answer














                                                                              share|cite|improve this answer



                                                                              share|cite|improve this answer








                                                                              edited Jul 29 '14 at 5:41









                                                                              Tunk-Fey

                                                                              22.9k868100




                                                                              22.9k868100










                                                                              answered Nov 1 '13 at 17:02









                                                                              Khosrotash

                                                                              16.8k12361




                                                                              16.8k12361








                                                                              • 11




                                                                                Ideally use Latex.
                                                                                – Meow
                                                                                Nov 1 '13 at 17:13








                                                                              • 2




                                                                                I am afraid that this approach is incorrect, since similar versions of it can be applied to convergent infinite series. The reason that it does not work is because the number of terms is infinite. All you've proven is that the second series approaches the final value faster than the first one. But whether this value is ultimately finite or not, you have not shown.
                                                                                – Lucian
                                                                                Jan 14 '15 at 16:32
















                                                                              • 11




                                                                                Ideally use Latex.
                                                                                – Meow
                                                                                Nov 1 '13 at 17:13








                                                                              • 2




                                                                                I am afraid that this approach is incorrect, since similar versions of it can be applied to convergent infinite series. The reason that it does not work is because the number of terms is infinite. All you've proven is that the second series approaches the final value faster than the first one. But whether this value is ultimately finite or not, you have not shown.
                                                                                – Lucian
                                                                                Jan 14 '15 at 16:32










                                                                              11




                                                                              11




                                                                              Ideally use Latex.
                                                                              – Meow
                                                                              Nov 1 '13 at 17:13






                                                                              Ideally use Latex.
                                                                              – Meow
                                                                              Nov 1 '13 at 17:13






                                                                              2




                                                                              2




                                                                              I am afraid that this approach is incorrect, since similar versions of it can be applied to convergent infinite series. The reason that it does not work is because the number of terms is infinite. All you've proven is that the second series approaches the final value faster than the first one. But whether this value is ultimately finite or not, you have not shown.
                                                                              – Lucian
                                                                              Jan 14 '15 at 16:32






                                                                              I am afraid that this approach is incorrect, since similar versions of it can be applied to convergent infinite series. The reason that it does not work is because the number of terms is infinite. All you've proven is that the second series approaches the final value faster than the first one. But whether this value is ultimately finite or not, you have not shown.
                                                                              – Lucian
                                                                              Jan 14 '15 at 16:32












                                                                              up vote
                                                                              6
                                                                              down vote













                                                                              A non-rigorous explanation I thought of once: consider a savings scheme where you put a dollar in your piggy bank every day. So after $n$ days, you have $n$ dollars; clearly, your savings approach infinity. On the other hand, each day you add an additional $1/n$ proportion of your existing savings, "so" (the non-rigorous step) the accumulated percentage after $n$ days is $1 + 1/2 + cdots + 1/n$.



                                                                              This can be made rigorous through the infinite product argument
                                                                              $$prod_{n = 1}^infty (1 + tfrac{1}{n}) < infty iff sum_{n = 1}^infty frac{1}{n} < infty$$
                                                                              which is obtained, essentially, by taking the logarithm of the left-hand side and using the power series for $log (1 + x)$.






                                                                              share|cite|improve this answer

























                                                                                up vote
                                                                                6
                                                                                down vote













                                                                                A non-rigorous explanation I thought of once: consider a savings scheme where you put a dollar in your piggy bank every day. So after $n$ days, you have $n$ dollars; clearly, your savings approach infinity. On the other hand, each day you add an additional $1/n$ proportion of your existing savings, "so" (the non-rigorous step) the accumulated percentage after $n$ days is $1 + 1/2 + cdots + 1/n$.



                                                                                This can be made rigorous through the infinite product argument
                                                                                $$prod_{n = 1}^infty (1 + tfrac{1}{n}) < infty iff sum_{n = 1}^infty frac{1}{n} < infty$$
                                                                                which is obtained, essentially, by taking the logarithm of the left-hand side and using the power series for $log (1 + x)$.






                                                                                share|cite|improve this answer























                                                                                  up vote
                                                                                  6
                                                                                  down vote










                                                                                  up vote
                                                                                  6
                                                                                  down vote









                                                                                  A non-rigorous explanation I thought of once: consider a savings scheme where you put a dollar in your piggy bank every day. So after $n$ days, you have $n$ dollars; clearly, your savings approach infinity. On the other hand, each day you add an additional $1/n$ proportion of your existing savings, "so" (the non-rigorous step) the accumulated percentage after $n$ days is $1 + 1/2 + cdots + 1/n$.



                                                                                  This can be made rigorous through the infinite product argument
                                                                                  $$prod_{n = 1}^infty (1 + tfrac{1}{n}) < infty iff sum_{n = 1}^infty frac{1}{n} < infty$$
                                                                                  which is obtained, essentially, by taking the logarithm of the left-hand side and using the power series for $log (1 + x)$.






                                                                                  share|cite|improve this answer












                                                                                  A non-rigorous explanation I thought of once: consider a savings scheme where you put a dollar in your piggy bank every day. So after $n$ days, you have $n$ dollars; clearly, your savings approach infinity. On the other hand, each day you add an additional $1/n$ proportion of your existing savings, "so" (the non-rigorous step) the accumulated percentage after $n$ days is $1 + 1/2 + cdots + 1/n$.



                                                                                  This can be made rigorous through the infinite product argument
                                                                                  $$prod_{n = 1}^infty (1 + tfrac{1}{n}) < infty iff sum_{n = 1}^infty frac{1}{n} < infty$$
                                                                                  which is obtained, essentially, by taking the logarithm of the left-hand side and using the power series for $log (1 + x)$.







                                                                                  share|cite|improve this answer












                                                                                  share|cite|improve this answer



                                                                                  share|cite|improve this answer










                                                                                  answered Nov 3 '13 at 18:37









                                                                                  Ryan Reich

                                                                                  5,3711627




                                                                                  5,3711627






















                                                                                      up vote
                                                                                      6
                                                                                      down vote













                                                                                      Another answer that's very similar to others. But it's prettier, and perhaps easier to understand for the 9-th grade student who asked the same question here.



                                                                                      The student's question was ... does the sum equal some number $S$. But, look:



                                                                                      enter image description here



                                                                                      So, whatever it is, $S$ is larger than the sum of the infinite string of $tfrac12$'s shown in the last line. No number can be this large, so $S$ can't be equal to any number. Mathematicians say that the series "diverges to infinity".






                                                                                      share|cite|improve this answer



























                                                                                        up vote
                                                                                        6
                                                                                        down vote













                                                                                        Another answer that's very similar to others. But it's prettier, and perhaps easier to understand for the 9-th grade student who asked the same question here.



                                                                                        The student's question was ... does the sum equal some number $S$. But, look:



                                                                                        enter image description here



                                                                                        So, whatever it is, $S$ is larger than the sum of the infinite string of $tfrac12$'s shown in the last line. No number can be this large, so $S$ can't be equal to any number. Mathematicians say that the series "diverges to infinity".






                                                                                        share|cite|improve this answer

























                                                                                          up vote
                                                                                          6
                                                                                          down vote










                                                                                          up vote
                                                                                          6
                                                                                          down vote









                                                                                          Another answer that's very similar to others. But it's prettier, and perhaps easier to understand for the 9-th grade student who asked the same question here.



                                                                                          The student's question was ... does the sum equal some number $S$. But, look:



                                                                                          enter image description here



                                                                                          So, whatever it is, $S$ is larger than the sum of the infinite string of $tfrac12$'s shown in the last line. No number can be this large, so $S$ can't be equal to any number. Mathematicians say that the series "diverges to infinity".






                                                                                          share|cite|improve this answer














                                                                                          Another answer that's very similar to others. But it's prettier, and perhaps easier to understand for the 9-th grade student who asked the same question here.



                                                                                          The student's question was ... does the sum equal some number $S$. But, look:



                                                                                          enter image description here



                                                                                          So, whatever it is, $S$ is larger than the sum of the infinite string of $tfrac12$'s shown in the last line. No number can be this large, so $S$ can't be equal to any number. Mathematicians say that the series "diverges to infinity".







                                                                                          share|cite|improve this answer














                                                                                          share|cite|improve this answer



                                                                                          share|cite|improve this answer








                                                                                          edited Apr 13 '17 at 12:21









                                                                                          Community

                                                                                          1




                                                                                          1










                                                                                          answered Sep 25 '16 at 2:17









                                                                                          bubba

                                                                                          29.5k32984




                                                                                          29.5k32984






















                                                                                              up vote
                                                                                              5
                                                                                              down vote













                                                                                              I think the integral test gives the most intuitive explanation. Observe that $$int^n_1 frac1x dx= log n$$ The sum $displaystylesum^n_{k=1}frac1k$ can be viewed as the area of $n$ rectangles of height $frac1k$, width $1$ (with the first one having it's left hand side on the y axis, and all having their bottom on the x axis). The graph of $xmapsto frac1x$ can be drawn under these, so the sum will grow with $n$ at (least) as fast as the integral - hence will grow (at least) logarithmically.






                                                                                              share|cite|improve this answer























                                                                                              • use log to get nice formatting for $log$
                                                                                                – Tyler
                                                                                                Nov 1 '13 at 17:24















                                                                                              up vote
                                                                                              5
                                                                                              down vote













                                                                                              I think the integral test gives the most intuitive explanation. Observe that $$int^n_1 frac1x dx= log n$$ The sum $displaystylesum^n_{k=1}frac1k$ can be viewed as the area of $n$ rectangles of height $frac1k$, width $1$ (with the first one having it's left hand side on the y axis, and all having their bottom on the x axis). The graph of $xmapsto frac1x$ can be drawn under these, so the sum will grow with $n$ at (least) as fast as the integral - hence will grow (at least) logarithmically.






                                                                                              share|cite|improve this answer























                                                                                              • use log to get nice formatting for $log$
                                                                                                – Tyler
                                                                                                Nov 1 '13 at 17:24













                                                                                              up vote
                                                                                              5
                                                                                              down vote










                                                                                              up vote
                                                                                              5
                                                                                              down vote









                                                                                              I think the integral test gives the most intuitive explanation. Observe that $$int^n_1 frac1x dx= log n$$ The sum $displaystylesum^n_{k=1}frac1k$ can be viewed as the area of $n$ rectangles of height $frac1k$, width $1$ (with the first one having it's left hand side on the y axis, and all having their bottom on the x axis). The graph of $xmapsto frac1x$ can be drawn under these, so the sum will grow with $n$ at (least) as fast as the integral - hence will grow (at least) logarithmically.






                                                                                              share|cite|improve this answer














                                                                                              I think the integral test gives the most intuitive explanation. Observe that $$int^n_1 frac1x dx= log n$$ The sum $displaystylesum^n_{k=1}frac1k$ can be viewed as the area of $n$ rectangles of height $frac1k$, width $1$ (with the first one having it's left hand side on the y axis, and all having their bottom on the x axis). The graph of $xmapsto frac1x$ can be drawn under these, so the sum will grow with $n$ at (least) as fast as the integral - hence will grow (at least) logarithmically.







                                                                                              share|cite|improve this answer














                                                                                              share|cite|improve this answer



                                                                                              share|cite|improve this answer








                                                                                              edited Jul 29 '14 at 5:43









                                                                                              Tunk-Fey

                                                                                              22.9k868100




                                                                                              22.9k868100










                                                                                              answered Nov 1 '13 at 17:09









                                                                                              Matt Rigby

                                                                                              2,085713




                                                                                              2,085713












                                                                                              • use log to get nice formatting for $log$
                                                                                                – Tyler
                                                                                                Nov 1 '13 at 17:24


















                                                                                              • use log to get nice formatting for $log$
                                                                                                – Tyler
                                                                                                Nov 1 '13 at 17:24
















                                                                                              use log to get nice formatting for $log$
                                                                                              – Tyler
                                                                                              Nov 1 '13 at 17:24




                                                                                              use log to get nice formatting for $log$
                                                                                              – Tyler
                                                                                              Nov 1 '13 at 17:24










                                                                                              up vote
                                                                                              4
                                                                                              down vote













                                                                                              Let be the partial sum $H_n = frac11 + frac12 + frac13 + cdots + frac1n$. Using Cesàro-Stolz:
                                                                                              $$
                                                                                              lim_{ntoinfty}frac{H_n}{log n} = lim_{ntoinfty}frac{H_{n+1}-H_n}{log(n+1)-log n} = lim_{ntoinfty}frac{frac1{n+1}}{log(1+1/n)}
                                                                                              = lim_{ntoinfty}frac{frac1{n+1}}{frac1n} = 1
                                                                                              $$
                                                                                              and
                                                                                              $$sum_{n=1}^inftyfrac1n = lim_{ntoinfty}H_n = infty.$$






                                                                                              share|cite|improve this answer

























                                                                                                up vote
                                                                                                4
                                                                                                down vote













                                                                                                Let be the partial sum $H_n = frac11 + frac12 + frac13 + cdots + frac1n$. Using Cesàro-Stolz:
                                                                                                $$
                                                                                                lim_{ntoinfty}frac{H_n}{log n} = lim_{ntoinfty}frac{H_{n+1}-H_n}{log(n+1)-log n} = lim_{ntoinfty}frac{frac1{n+1}}{log(1+1/n)}
                                                                                                = lim_{ntoinfty}frac{frac1{n+1}}{frac1n} = 1
                                                                                                $$
                                                                                                and
                                                                                                $$sum_{n=1}^inftyfrac1n = lim_{ntoinfty}H_n = infty.$$






                                                                                                share|cite|improve this answer























                                                                                                  up vote
                                                                                                  4
                                                                                                  down vote










                                                                                                  up vote
                                                                                                  4
                                                                                                  down vote









                                                                                                  Let be the partial sum $H_n = frac11 + frac12 + frac13 + cdots + frac1n$. Using Cesàro-Stolz:
                                                                                                  $$
                                                                                                  lim_{ntoinfty}frac{H_n}{log n} = lim_{ntoinfty}frac{H_{n+1}-H_n}{log(n+1)-log n} = lim_{ntoinfty}frac{frac1{n+1}}{log(1+1/n)}
                                                                                                  = lim_{ntoinfty}frac{frac1{n+1}}{frac1n} = 1
                                                                                                  $$
                                                                                                  and
                                                                                                  $$sum_{n=1}^inftyfrac1n = lim_{ntoinfty}H_n = infty.$$






                                                                                                  share|cite|improve this answer












                                                                                                  Let be the partial sum $H_n = frac11 + frac12 + frac13 + cdots + frac1n$. Using Cesàro-Stolz:
                                                                                                  $$
                                                                                                  lim_{ntoinfty}frac{H_n}{log n} = lim_{ntoinfty}frac{H_{n+1}-H_n}{log(n+1)-log n} = lim_{ntoinfty}frac{frac1{n+1}}{log(1+1/n)}
                                                                                                  = lim_{ntoinfty}frac{frac1{n+1}}{frac1n} = 1
                                                                                                  $$
                                                                                                  and
                                                                                                  $$sum_{n=1}^inftyfrac1n = lim_{ntoinfty}H_n = infty.$$







                                                                                                  share|cite|improve this answer












                                                                                                  share|cite|improve this answer



                                                                                                  share|cite|improve this answer










                                                                                                  answered Jan 30 '15 at 12:37









                                                                                                  Martín-Blas Pérez Pinilla

                                                                                                  33.7k42770




                                                                                                  33.7k42770






















                                                                                                      up vote
                                                                                                      2
                                                                                                      down vote













                                                                                                      We all know that $$sum_{n=1}^inftyfrac1n=frac 1 1 + frac 12 + frac 13 + cdots $$ diverges and grows very slowly!! I have seen many proofs of the result but recently found the following: $$S =frac 1 1 + frac 12 + frac 13 +frac 14+ frac 15+ frac 16+ cdots$$ $$> frac 12+frac 12+ frac 14+ frac 14+ frac 16+ frac 16+ cdots =frac 1 1 + frac 12 + frac 13 +cdots = S.$$
                                                                                                      In this way we see that $S > S$.






                                                                                                      share|cite|improve this answer





















                                                                                                      • O.o This. Is. Amazing!! =)
                                                                                                        – user378947
                                                                                                        Dec 12 '16 at 1:18










                                                                                                      • You can also see it here math.stackexchange.com/questions/1160527/…
                                                                                                        – user8795
                                                                                                        Dec 12 '16 at 1:20










                                                                                                      • I have saved this to my personal The Book :) That being said... Come on! The last inequality itself is proof enough!! :P
                                                                                                        – user378947
                                                                                                        Dec 12 '16 at 1:31















                                                                                                      up vote
                                                                                                      2
                                                                                                      down vote













                                                                                                      We all know that $$sum_{n=1}^inftyfrac1n=frac 1 1 + frac 12 + frac 13 + cdots $$ diverges and grows very slowly!! I have seen many proofs of the result but recently found the following: $$S =frac 1 1 + frac 12 + frac 13 +frac 14+ frac 15+ frac 16+ cdots$$ $$> frac 12+frac 12+ frac 14+ frac 14+ frac 16+ frac 16+ cdots =frac 1 1 + frac 12 + frac 13 +cdots = S.$$
                                                                                                      In this way we see that $S > S$.






                                                                                                      share|cite|improve this answer





















                                                                                                      • O.o This. Is. Amazing!! =)
                                                                                                        – user378947
                                                                                                        Dec 12 '16 at 1:18










                                                                                                      • You can also see it here math.stackexchange.com/questions/1160527/…
                                                                                                        – user8795
                                                                                                        Dec 12 '16 at 1:20










                                                                                                      • I have saved this to my personal The Book :) That being said... Come on! The last inequality itself is proof enough!! :P
                                                                                                        – user378947
                                                                                                        Dec 12 '16 at 1:31













                                                                                                      up vote
                                                                                                      2
                                                                                                      down vote










                                                                                                      up vote
                                                                                                      2
                                                                                                      down vote









                                                                                                      We all know that $$sum_{n=1}^inftyfrac1n=frac 1 1 + frac 12 + frac 13 + cdots $$ diverges and grows very slowly!! I have seen many proofs of the result but recently found the following: $$S =frac 1 1 + frac 12 + frac 13 +frac 14+ frac 15+ frac 16+ cdots$$ $$> frac 12+frac 12+ frac 14+ frac 14+ frac 16+ frac 16+ cdots =frac 1 1 + frac 12 + frac 13 +cdots = S.$$
                                                                                                      In this way we see that $S > S$.






                                                                                                      share|cite|improve this answer












                                                                                                      We all know that $$sum_{n=1}^inftyfrac1n=frac 1 1 + frac 12 + frac 13 + cdots $$ diverges and grows very slowly!! I have seen many proofs of the result but recently found the following: $$S =frac 1 1 + frac 12 + frac 13 +frac 14+ frac 15+ frac 16+ cdots$$ $$> frac 12+frac 12+ frac 14+ frac 14+ frac 16+ frac 16+ cdots =frac 1 1 + frac 12 + frac 13 +cdots = S.$$
                                                                                                      In this way we see that $S > S$.







                                                                                                      share|cite|improve this answer












                                                                                                      share|cite|improve this answer



                                                                                                      share|cite|improve this answer










                                                                                                      answered Feb 22 '15 at 19:25









                                                                                                      user8795

                                                                                                      5,56461846




                                                                                                      5,56461846












                                                                                                      • O.o This. Is. Amazing!! =)
                                                                                                        – user378947
                                                                                                        Dec 12 '16 at 1:18










                                                                                                      • You can also see it here math.stackexchange.com/questions/1160527/…
                                                                                                        – user8795
                                                                                                        Dec 12 '16 at 1:20










                                                                                                      • I have saved this to my personal The Book :) That being said... Come on! The last inequality itself is proof enough!! :P
                                                                                                        – user378947
                                                                                                        Dec 12 '16 at 1:31


















                                                                                                      • O.o This. Is. Amazing!! =)
                                                                                                        – user378947
                                                                                                        Dec 12 '16 at 1:18










                                                                                                      • You can also see it here math.stackexchange.com/questions/1160527/…
                                                                                                        – user8795
                                                                                                        Dec 12 '16 at 1:20










                                                                                                      • I have saved this to my personal The Book :) That being said... Come on! The last inequality itself is proof enough!! :P
                                                                                                        – user378947
                                                                                                        Dec 12 '16 at 1:31
















                                                                                                      O.o This. Is. Amazing!! =)
                                                                                                      – user378947
                                                                                                      Dec 12 '16 at 1:18




                                                                                                      O.o This. Is. Amazing!! =)
                                                                                                      – user378947
                                                                                                      Dec 12 '16 at 1:18












                                                                                                      You can also see it here math.stackexchange.com/questions/1160527/…
                                                                                                      – user8795
                                                                                                      Dec 12 '16 at 1:20




                                                                                                      You can also see it here math.stackexchange.com/questions/1160527/…
                                                                                                      – user8795
                                                                                                      Dec 12 '16 at 1:20












                                                                                                      I have saved this to my personal The Book :) That being said... Come on! The last inequality itself is proof enough!! :P
                                                                                                      – user378947
                                                                                                      Dec 12 '16 at 1:31




                                                                                                      I have saved this to my personal The Book :) That being said... Come on! The last inequality itself is proof enough!! :P
                                                                                                      – user378947
                                                                                                      Dec 12 '16 at 1:31










                                                                                                      up vote
                                                                                                      2
                                                                                                      down vote













                                                                                                      Using Euler's form of the Harmonic numbers,



                                                                                                      $$sum_{k=1}^nfrac1k=int_0^1frac{1-x^n}{1-x}dx$$



                                                                                                      $$begin{align}
                                                                                                      lim_{ntoinfty}sum_{k=1}^nfrac1k & =lim_{ntoinfty}int_0^1frac{1-x^n}{1-x}dx \
                                                                                                      & =int_0^1frac1{1-x}dx \
                                                                                                      & =left.lim_{pto1^+}-ln(1-x)right]_0^p \
                                                                                                      & to+infty
                                                                                                      end{align}$$





                                                                                                      Using the Taylor expansion of $ln(1-x)$,



                                                                                                      $$-ln(1-x)=x+frac{x^2}2+frac{x^3}3+frac{x^4}4+dots$$



                                                                                                      $$-ln(1-1)=1+frac12+frac13+frac14+dotsquad $$





                                                                                                      Using Euler's relationship between the Riemann zeta function and the Dirichlet eta function,



                                                                                                      $$begin{align}
                                                                                                      sum_{k=1}^inftyfrac1{k^s} & =frac1{1-2^{1-s}}sum_{k=1}^inftyfrac{(-1)^{k+1}}{k^s} \
                                                                                                      sum_{k=1}^inftyfrac1k & =frac10sum_{k=1}^inftyfrac{(-1)^{k+1}}ktag{$s=1$} \
                                                                                                      & to+infty
                                                                                                      end{align}$$






                                                                                                      share|cite|improve this answer























                                                                                                      • But isn't the series for $ln(1-x)$ only valid for $-1le x<1$?
                                                                                                        – TheSimpliFire
                                                                                                        Mar 17 at 9:31












                                                                                                      • Yes, but since the limit as $xto1$ in $x^n$ is monotone, it equals the asked series, if they exist.
                                                                                                        – Simply Beautiful Art
                                                                                                        Mar 19 at 2:33















                                                                                                      up vote
                                                                                                      2
                                                                                                      down vote













                                                                                                      Using Euler's form of the Harmonic numbers,



                                                                                                      $$sum_{k=1}^nfrac1k=int_0^1frac{1-x^n}{1-x}dx$$



                                                                                                      $$begin{align}
                                                                                                      lim_{ntoinfty}sum_{k=1}^nfrac1k & =lim_{ntoinfty}int_0^1frac{1-x^n}{1-x}dx \
                                                                                                      & =int_0^1frac1{1-x}dx \
                                                                                                      & =left.lim_{pto1^+}-ln(1-x)right]_0^p \
                                                                                                      & to+infty
                                                                                                      end{align}$$





                                                                                                      Using the Taylor expansion of $ln(1-x)$,



                                                                                                      $$-ln(1-x)=x+frac{x^2}2+frac{x^3}3+frac{x^4}4+dots$$



                                                                                                      $$-ln(1-1)=1+frac12+frac13+frac14+dotsquad $$





                                                                                                      Using Euler's relationship between the Riemann zeta function and the Dirichlet eta function,



                                                                                                      $$begin{align}
                                                                                                      sum_{k=1}^inftyfrac1{k^s} & =frac1{1-2^{1-s}}sum_{k=1}^inftyfrac{(-1)^{k+1}}{k^s} \
                                                                                                      sum_{k=1}^inftyfrac1k & =frac10sum_{k=1}^inftyfrac{(-1)^{k+1}}ktag{$s=1$} \
                                                                                                      & to+infty
                                                                                                      end{align}$$






                                                                                                      share|cite|improve this answer























                                                                                                      • But isn't the series for $ln(1-x)$ only valid for $-1le x<1$?
                                                                                                        – TheSimpliFire
                                                                                                        Mar 17 at 9:31












                                                                                                      • Yes, but since the limit as $xto1$ in $x^n$ is monotone, it equals the asked series, if they exist.
                                                                                                        – Simply Beautiful Art
                                                                                                        Mar 19 at 2:33













                                                                                                      up vote
                                                                                                      2
                                                                                                      down vote










                                                                                                      up vote
                                                                                                      2
                                                                                                      down vote









                                                                                                      Using Euler's form of the Harmonic numbers,



                                                                                                      $$sum_{k=1}^nfrac1k=int_0^1frac{1-x^n}{1-x}dx$$



                                                                                                      $$begin{align}
                                                                                                      lim_{ntoinfty}sum_{k=1}^nfrac1k & =lim_{ntoinfty}int_0^1frac{1-x^n}{1-x}dx \
                                                                                                      & =int_0^1frac1{1-x}dx \
                                                                                                      & =left.lim_{pto1^+}-ln(1-x)right]_0^p \
                                                                                                      & to+infty
                                                                                                      end{align}$$





                                                                                                      Using the Taylor expansion of $ln(1-x)$,



                                                                                                      $$-ln(1-x)=x+frac{x^2}2+frac{x^3}3+frac{x^4}4+dots$$



                                                                                                      $$-ln(1-1)=1+frac12+frac13+frac14+dotsquad $$





                                                                                                      Using Euler's relationship between the Riemann zeta function and the Dirichlet eta function,



                                                                                                      $$begin{align}
                                                                                                      sum_{k=1}^inftyfrac1{k^s} & =frac1{1-2^{1-s}}sum_{k=1}^inftyfrac{(-1)^{k+1}}{k^s} \
                                                                                                      sum_{k=1}^inftyfrac1k & =frac10sum_{k=1}^inftyfrac{(-1)^{k+1}}ktag{$s=1$} \
                                                                                                      & to+infty
                                                                                                      end{align}$$






                                                                                                      share|cite|improve this answer














                                                                                                      Using Euler's form of the Harmonic numbers,



                                                                                                      $$sum_{k=1}^nfrac1k=int_0^1frac{1-x^n}{1-x}dx$$



                                                                                                      $$begin{align}
                                                                                                      lim_{ntoinfty}sum_{k=1}^nfrac1k & =lim_{ntoinfty}int_0^1frac{1-x^n}{1-x}dx \
                                                                                                      & =int_0^1frac1{1-x}dx \
                                                                                                      & =left.lim_{pto1^+}-ln(1-x)right]_0^p \
                                                                                                      & to+infty
                                                                                                      end{align}$$





                                                                                                      Using the Taylor expansion of $ln(1-x)$,



                                                                                                      $$-ln(1-x)=x+frac{x^2}2+frac{x^3}3+frac{x^4}4+dots$$



                                                                                                      $$-ln(1-1)=1+frac12+frac13+frac14+dotsquad $$





                                                                                                      Using Euler's relationship between the Riemann zeta function and the Dirichlet eta function,



                                                                                                      $$begin{align}
                                                                                                      sum_{k=1}^inftyfrac1{k^s} & =frac1{1-2^{1-s}}sum_{k=1}^inftyfrac{(-1)^{k+1}}{k^s} \
                                                                                                      sum_{k=1}^inftyfrac1k & =frac10sum_{k=1}^inftyfrac{(-1)^{k+1}}ktag{$s=1$} \
                                                                                                      & to+infty
                                                                                                      end{align}$$







                                                                                                      share|cite|improve this answer














                                                                                                      share|cite|improve this answer



                                                                                                      share|cite|improve this answer








                                                                                                      edited Oct 6 '16 at 17:16

























                                                                                                      answered Oct 6 '16 at 16:57









                                                                                                      Simply Beautiful Art

                                                                                                      50k577181




                                                                                                      50k577181












                                                                                                      • But isn't the series for $ln(1-x)$ only valid for $-1le x<1$?
                                                                                                        – TheSimpliFire
                                                                                                        Mar 17 at 9:31












                                                                                                      • Yes, but since the limit as $xto1$ in $x^n$ is monotone, it equals the asked series, if they exist.
                                                                                                        – Simply Beautiful Art
                                                                                                        Mar 19 at 2:33


















                                                                                                      • But isn't the series for $ln(1-x)$ only valid for $-1le x<1$?
                                                                                                        – TheSimpliFire
                                                                                                        Mar 17 at 9:31












                                                                                                      • Yes, but since the limit as $xto1$ in $x^n$ is monotone, it equals the asked series, if they exist.
                                                                                                        – Simply Beautiful Art
                                                                                                        Mar 19 at 2:33
















                                                                                                      But isn't the series for $ln(1-x)$ only valid for $-1le x<1$?
                                                                                                      – TheSimpliFire
                                                                                                      Mar 17 at 9:31






                                                                                                      But isn't the series for $ln(1-x)$ only valid for $-1le x<1$?
                                                                                                      – TheSimpliFire
                                                                                                      Mar 17 at 9:31














                                                                                                      Yes, but since the limit as $xto1$ in $x^n$ is monotone, it equals the asked series, if they exist.
                                                                                                      – Simply Beautiful Art
                                                                                                      Mar 19 at 2:33




                                                                                                      Yes, but since the limit as $xto1$ in $x^n$ is monotone, it equals the asked series, if they exist.
                                                                                                      – Simply Beautiful Art
                                                                                                      Mar 19 at 2:33










                                                                                                      up vote
                                                                                                      -3
                                                                                                      down vote













                                                                                                      A series converges if and only if the tail of the series tends to zero, i.e. the summation from N to infinity tends to zero for N to infinity. But in case of the Harmonic series, we have that the summation from N to 2 N is larger than the smallest term ( which is 1/(2N)), times the number of terms N, which yields 1/2. So, the tail clearly does not tend to zero for N to infinity.






                                                                                                      share|cite|improve this answer



















                                                                                                      • 2




                                                                                                        What do you think you were adding that hasn't been addressed thoroughly?
                                                                                                        – user223391
                                                                                                        Jan 31 '16 at 1:05










                                                                                                      • @avid19 I added an argument that's tl;dr-proof.
                                                                                                        – Count Iblis
                                                                                                        Jan 31 '16 at 4:29










                                                                                                      • @ZacharySelk My proof is the best, as it's the most concise, self contained proof given. Unlike the other proofs my proof can be modified into an argument that a 6 year old could understand (with some effort).
                                                                                                        – Count Iblis
                                                                                                        Oct 6 '16 at 22:35






                                                                                                      • 2




                                                                                                        @CountIblis Your proof is mathematically identical to the six-year-earlier (and top-voted, and accepted) answer by AgCl, which also explains the situation much more clearly. (Note that bringing tails into the picture is unnecessary, and just adds a layer of complexity: we can reason about the blocks directly as AgCl does, and as you do essentially, to show that the harmonic series goes to infinity.)
                                                                                                        – Noah Schweber
                                                                                                        Oct 11 '16 at 1:52

















                                                                                                      up vote
                                                                                                      -3
                                                                                                      down vote













                                                                                                      A series converges if and only if the tail of the series tends to zero, i.e. the summation from N to infinity tends to zero for N to infinity. But in case of the Harmonic series, we have that the summation from N to 2 N is larger than the smallest term ( which is 1/(2N)), times the number of terms N, which yields 1/2. So, the tail clearly does not tend to zero for N to infinity.






                                                                                                      share|cite|improve this answer



















                                                                                                      • 2




                                                                                                        What do you think you were adding that hasn't been addressed thoroughly?
                                                                                                        – user223391
                                                                                                        Jan 31 '16 at 1:05










                                                                                                      • @avid19 I added an argument that's tl;dr-proof.
                                                                                                        – Count Iblis
                                                                                                        Jan 31 '16 at 4:29










                                                                                                      • @ZacharySelk My proof is the best, as it's the most concise, self contained proof given. Unlike the other proofs my proof can be modified into an argument that a 6 year old could understand (with some effort).
                                                                                                        – Count Iblis
                                                                                                        Oct 6 '16 at 22:35






                                                                                                      • 2




                                                                                                        @CountIblis Your proof is mathematically identical to the six-year-earlier (and top-voted, and accepted) answer by AgCl, which also explains the situation much more clearly. (Note that bringing tails into the picture is unnecessary, and just adds a layer of complexity: we can reason about the blocks directly as AgCl does, and as you do essentially, to show that the harmonic series goes to infinity.)
                                                                                                        – Noah Schweber
                                                                                                        Oct 11 '16 at 1:52















                                                                                                      up vote
                                                                                                      -3
                                                                                                      down vote










                                                                                                      up vote
                                                                                                      -3
                                                                                                      down vote









                                                                                                      A series converges if and only if the tail of the series tends to zero, i.e. the summation from N to infinity tends to zero for N to infinity. But in case of the Harmonic series, we have that the summation from N to 2 N is larger than the smallest term ( which is 1/(2N)), times the number of terms N, which yields 1/2. So, the tail clearly does not tend to zero for N to infinity.






                                                                                                      share|cite|improve this answer














                                                                                                      A series converges if and only if the tail of the series tends to zero, i.e. the summation from N to infinity tends to zero for N to infinity. But in case of the Harmonic series, we have that the summation from N to 2 N is larger than the smallest term ( which is 1/(2N)), times the number of terms N, which yields 1/2. So, the tail clearly does not tend to zero for N to infinity.







                                                                                                      share|cite|improve this answer














                                                                                                      share|cite|improve this answer



                                                                                                      share|cite|improve this answer








                                                                                                      edited Oct 6 '16 at 22:29

























                                                                                                      answered Jan 31 '16 at 0:07









                                                                                                      Count Iblis

                                                                                                      8,15121333




                                                                                                      8,15121333








                                                                                                      • 2




                                                                                                        What do you think you were adding that hasn't been addressed thoroughly?
                                                                                                        – user223391
                                                                                                        Jan 31 '16 at 1:05










                                                                                                      • @avid19 I added an argument that's tl;dr-proof.
                                                                                                        – Count Iblis
                                                                                                        Jan 31 '16 at 4:29










                                                                                                      • @ZacharySelk My proof is the best, as it's the most concise, self contained proof given. Unlike the other proofs my proof can be modified into an argument that a 6 year old could understand (with some effort).
                                                                                                        – Count Iblis
                                                                                                        Oct 6 '16 at 22:35






                                                                                                      • 2




                                                                                                        @CountIblis Your proof is mathematically identical to the six-year-earlier (and top-voted, and accepted) answer by AgCl, which also explains the situation much more clearly. (Note that bringing tails into the picture is unnecessary, and just adds a layer of complexity: we can reason about the blocks directly as AgCl does, and as you do essentially, to show that the harmonic series goes to infinity.)
                                                                                                        – Noah Schweber
                                                                                                        Oct 11 '16 at 1:52
















                                                                                                      • 2




                                                                                                        What do you think you were adding that hasn't been addressed thoroughly?
                                                                                                        – user223391
                                                                                                        Jan 31 '16 at 1:05










                                                                                                      • @avid19 I added an argument that's tl;dr-proof.
                                                                                                        – Count Iblis
                                                                                                        Jan 31 '16 at 4:29










                                                                                                      • @ZacharySelk My proof is the best, as it's the most concise, self contained proof given. Unlike the other proofs my proof can be modified into an argument that a 6 year old could understand (with some effort).
                                                                                                        – Count Iblis
                                                                                                        Oct 6 '16 at 22:35






                                                                                                      • 2




                                                                                                        @CountIblis Your proof is mathematically identical to the six-year-earlier (and top-voted, and accepted) answer by AgCl, which also explains the situation much more clearly. (Note that bringing tails into the picture is unnecessary, and just adds a layer of complexity: we can reason about the blocks directly as AgCl does, and as you do essentially, to show that the harmonic series goes to infinity.)
                                                                                                        – Noah Schweber
                                                                                                        Oct 11 '16 at 1:52










                                                                                                      2




                                                                                                      2




                                                                                                      What do you think you were adding that hasn't been addressed thoroughly?
                                                                                                      – user223391
                                                                                                      Jan 31 '16 at 1:05




                                                                                                      What do you think you were adding that hasn't been addressed thoroughly?
                                                                                                      – user223391
                                                                                                      Jan 31 '16 at 1:05












                                                                                                      @avid19 I added an argument that's tl;dr-proof.
                                                                                                      – Count Iblis
                                                                                                      Jan 31 '16 at 4:29




                                                                                                      @avid19 I added an argument that's tl;dr-proof.
                                                                                                      – Count Iblis
                                                                                                      Jan 31 '16 at 4:29












                                                                                                      @ZacharySelk My proof is the best, as it's the most concise, self contained proof given. Unlike the other proofs my proof can be modified into an argument that a 6 year old could understand (with some effort).
                                                                                                      – Count Iblis
                                                                                                      Oct 6 '16 at 22:35




                                                                                                      @ZacharySelk My proof is the best, as it's the most concise, self contained proof given. Unlike the other proofs my proof can be modified into an argument that a 6 year old could understand (with some effort).
                                                                                                      – Count Iblis
                                                                                                      Oct 6 '16 at 22:35




                                                                                                      2




                                                                                                      2




                                                                                                      @CountIblis Your proof is mathematically identical to the six-year-earlier (and top-voted, and accepted) answer by AgCl, which also explains the situation much more clearly. (Note that bringing tails into the picture is unnecessary, and just adds a layer of complexity: we can reason about the blocks directly as AgCl does, and as you do essentially, to show that the harmonic series goes to infinity.)
                                                                                                      – Noah Schweber
                                                                                                      Oct 11 '16 at 1:52






                                                                                                      @CountIblis Your proof is mathematically identical to the six-year-earlier (and top-voted, and accepted) answer by AgCl, which also explains the situation much more clearly. (Note that bringing tails into the picture is unnecessary, and just adds a layer of complexity: we can reason about the blocks directly as AgCl does, and as you do essentially, to show that the harmonic series goes to infinity.)
                                                                                                      – Noah Schweber
                                                                                                      Oct 11 '16 at 1:52




















                                                                                                       

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