partial n-th sum of the infinity sequence and converging












-1














I am suppose to find partial n-th sum of the infinity sequence and determine if it converge.



My sequence is $$ sum_{ n=1}^{infty} ln left(1+frac{1}{n}right) $$



My solution is :



I find out that first terms are:$ ln (2), lnfrac{3}{2}, ln frac{4}{3} $



Then I wrote that: $ s_{n}=(ln(2)+ln(frac{3}{2})+lnfrac{4}{3}+...ln(1+frac{1}{n}) ) $



I know that, it is not enough, but I have no idea, what to do next, or if this is true.



Thanks a lot for any advice.






real-analysis







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    -1














    I am suppose to find partial n-th sum of the infinity sequence and determine if it converge.



    My sequence is $$ sum_{ n=1}^{infty} ln left(1+frac{1}{n}right) $$



    My solution is :



    I find out that first terms are:$ ln (2), lnfrac{3}{2}, ln frac{4}{3} $



    Then I wrote that: $ s_{n}=(ln(2)+ln(frac{3}{2})+lnfrac{4}{3}+...ln(1+frac{1}{n}) ) $



    I know that, it is not enough, but I have no idea, what to do next, or if this is true.



    Thanks a lot for any advice.






    real-analysis







    share|cite|improve this question



























      -1












      -1








      -1







      I am suppose to find partial n-th sum of the infinity sequence and determine if it converge.



      My sequence is $$ sum_{ n=1}^{infty} ln left(1+frac{1}{n}right) $$



      My solution is :



      I find out that first terms are:$ ln (2), lnfrac{3}{2}, ln frac{4}{3} $



      Then I wrote that: $ s_{n}=(ln(2)+ln(frac{3}{2})+lnfrac{4}{3}+...ln(1+frac{1}{n}) ) $



      I know that, it is not enough, but I have no idea, what to do next, or if this is true.



      Thanks a lot for any advice.






      real-analysis







      share|cite|improve this question















      I am suppose to find partial n-th sum of the infinity sequence and determine if it converge.



      My sequence is $$ sum_{ n=1}^{infty} ln left(1+frac{1}{n}right) $$



      My solution is :



      I find out that first terms are:$ ln (2), lnfrac{3}{2}, ln frac{4}{3} $



      Then I wrote that: $ s_{n}=(ln(2)+ln(frac{3}{2})+lnfrac{4}{3}+...ln(1+frac{1}{n}) ) $



      I know that, it is not enough, but I have no idea, what to do next, or if this is true.



      Thanks a lot for any advice.






      real-analysis




      real-analysis






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      share|cite|improve this question













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      edited Dec 2 at 17:10









      Larry

      1,6682822




      1,6682822










      asked Dec 2 at 16:39









      Shelley

      92




      92






















          3 Answers
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          $$sum_{k=1}^n ln(1+1/k)=sum_{k=1}^nlnleft(frac{k+1}{k}right) =lnprod_{k=1}^n frac{k+1}{k} =lnfrac{(n+1)!}{n!}=ln(n+1)$$






          share|cite|improve this answer





















          • I do not understand, why did you use factorial?
            – Shelley
            Dec 2 at 16:53










          • Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
            – Tito Eliatron
            Dec 2 at 16:56



















          0














          $$lnleft(frac{k+1}{k}right)=ln(k+1)-ln(k)$$
          $$sum_{k=1}^{n} ln left(1+frac{1}{k}right) = sum_{k=1}^{n}lnleft(frac{k+1}{k}right)=sum_{k=1}^{n}ln(k+1)-ln(k)=ln(n+1)$$






          share|cite|improve this answer





























            0














            As $xto 0,$



            $$frac{ln(1+x)}{x}= frac{ln(1+x)-ln 1}{x-0}to ln'(1)=1.$$



            This shows $[ln(1+1/n)]/(1/n)to 1.$ We know $sum 1/n$ diverges, so by the limit comparison test, $sum ln(1+1/n)$ diverges.






            share|cite|improve this answer





















            • More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
              – zhw.
              Dec 2 at 17:18













            Your Answer





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            3 Answers
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            active

            oldest

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            3 Answers
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            active

            oldest

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            active

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            active

            oldest

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            1














            $$sum_{k=1}^n ln(1+1/k)=sum_{k=1}^nlnleft(frac{k+1}{k}right) =lnprod_{k=1}^n frac{k+1}{k} =lnfrac{(n+1)!}{n!}=ln(n+1)$$






            share|cite|improve this answer





















            • I do not understand, why did you use factorial?
              – Shelley
              Dec 2 at 16:53










            • Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
              – Tito Eliatron
              Dec 2 at 16:56
















            1














            $$sum_{k=1}^n ln(1+1/k)=sum_{k=1}^nlnleft(frac{k+1}{k}right) =lnprod_{k=1}^n frac{k+1}{k} =lnfrac{(n+1)!}{n!}=ln(n+1)$$






            share|cite|improve this answer





















            • I do not understand, why did you use factorial?
              – Shelley
              Dec 2 at 16:53










            • Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
              – Tito Eliatron
              Dec 2 at 16:56














            1












            1








            1






            $$sum_{k=1}^n ln(1+1/k)=sum_{k=1}^nlnleft(frac{k+1}{k}right) =lnprod_{k=1}^n frac{k+1}{k} =lnfrac{(n+1)!}{n!}=ln(n+1)$$






            share|cite|improve this answer












            $$sum_{k=1}^n ln(1+1/k)=sum_{k=1}^nlnleft(frac{k+1}{k}right) =lnprod_{k=1}^n frac{k+1}{k} =lnfrac{(n+1)!}{n!}=ln(n+1)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 2 at 16:44









            Tito Eliatron

            1,553622




            1,553622












            • I do not understand, why did you use factorial?
              – Shelley
              Dec 2 at 16:53










            • Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
              – Tito Eliatron
              Dec 2 at 16:56


















            • I do not understand, why did you use factorial?
              – Shelley
              Dec 2 at 16:53










            • Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
              – Tito Eliatron
              Dec 2 at 16:56
















            I do not understand, why did you use factorial?
            – Shelley
            Dec 2 at 16:53




            I do not understand, why did you use factorial?
            – Shelley
            Dec 2 at 16:53












            Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
            – Tito Eliatron
            Dec 2 at 16:56




            Because is the product of natural numbers from $n+1$ to 2 (in the numerator); and from $n$ to $1$ in the denominator..
            – Tito Eliatron
            Dec 2 at 16:56











            0














            $$lnleft(frac{k+1}{k}right)=ln(k+1)-ln(k)$$
            $$sum_{k=1}^{n} ln left(1+frac{1}{k}right) = sum_{k=1}^{n}lnleft(frac{k+1}{k}right)=sum_{k=1}^{n}ln(k+1)-ln(k)=ln(n+1)$$






            share|cite|improve this answer


























              0














              $$lnleft(frac{k+1}{k}right)=ln(k+1)-ln(k)$$
              $$sum_{k=1}^{n} ln left(1+frac{1}{k}right) = sum_{k=1}^{n}lnleft(frac{k+1}{k}right)=sum_{k=1}^{n}ln(k+1)-ln(k)=ln(n+1)$$






              share|cite|improve this answer
























                0












                0








                0






                $$lnleft(frac{k+1}{k}right)=ln(k+1)-ln(k)$$
                $$sum_{k=1}^{n} ln left(1+frac{1}{k}right) = sum_{k=1}^{n}lnleft(frac{k+1}{k}right)=sum_{k=1}^{n}ln(k+1)-ln(k)=ln(n+1)$$






                share|cite|improve this answer












                $$lnleft(frac{k+1}{k}right)=ln(k+1)-ln(k)$$
                $$sum_{k=1}^{n} ln left(1+frac{1}{k}right) = sum_{k=1}^{n}lnleft(frac{k+1}{k}right)=sum_{k=1}^{n}ln(k+1)-ln(k)=ln(n+1)$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 at 16:57









                Larry

                1,6682822




                1,6682822























                    0














                    As $xto 0,$



                    $$frac{ln(1+x)}{x}= frac{ln(1+x)-ln 1}{x-0}to ln'(1)=1.$$



                    This shows $[ln(1+1/n)]/(1/n)to 1.$ We know $sum 1/n$ diverges, so by the limit comparison test, $sum ln(1+1/n)$ diverges.






                    share|cite|improve this answer





















                    • More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                      – zhw.
                      Dec 2 at 17:18


















                    0














                    As $xto 0,$



                    $$frac{ln(1+x)}{x}= frac{ln(1+x)-ln 1}{x-0}to ln'(1)=1.$$



                    This shows $[ln(1+1/n)]/(1/n)to 1.$ We know $sum 1/n$ diverges, so by the limit comparison test, $sum ln(1+1/n)$ diverges.






                    share|cite|improve this answer





















                    • More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                      – zhw.
                      Dec 2 at 17:18
















                    0












                    0








                    0






                    As $xto 0,$



                    $$frac{ln(1+x)}{x}= frac{ln(1+x)-ln 1}{x-0}to ln'(1)=1.$$



                    This shows $[ln(1+1/n)]/(1/n)to 1.$ We know $sum 1/n$ diverges, so by the limit comparison test, $sum ln(1+1/n)$ diverges.






                    share|cite|improve this answer












                    As $xto 0,$



                    $$frac{ln(1+x)}{x}= frac{ln(1+x)-ln 1}{x-0}to ln'(1)=1.$$



                    This shows $[ln(1+1/n)]/(1/n)to 1.$ We know $sum 1/n$ diverges, so by the limit comparison test, $sum ln(1+1/n)$ diverges.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 2 at 17:15









                    zhw.

                    71.6k43075




                    71.6k43075












                    • More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                      – zhw.
                      Dec 2 at 17:18




















                    • More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                      – zhw.
                      Dec 2 at 17:18


















                    More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                    – zhw.
                    Dec 2 at 17:18






                    More generally: if $f(0)=0$ and $f'(0)ne 0,$ then $sum f(1/n)$ diverges.
                    – zhw.
                    Dec 2 at 17:18




















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