What is an intuitive approach to solving $lim_{nrightarrowinfty}biggl(frac{1}{n^2} + frac{2}{n^2} +...











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$$lim_{nrightarrowinfty}biggl(frac{1}{n^2} + frac{2}{n^2} + frac{3}{n^2}+dots+frac{n}{n^2}biggr)$$




I managed to get the answer as $1$ by standard methods of solving, learned from teachers, but my intuition says that the denominator of every term grows much faster than the numerator so limit must equal to zero.



Where is my mistake? Please explain very intuitively.










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  • 14




    Human intuition in general is a product of evolution and history has shown that deep mathematical skills don't offer any extra advantages in human survival. Situation is similar to the fact that results of relativity and quantum mechanics are counter-intuitive. Nobody on earth had to walk with speeds close to light or deal with objects the size of an atom. Relying on intuition and trying to explain everything intuitively even at the expense of correctness is a modern trend and is based on the belief that it leads to better understanding. Cont'd..
    – Paramanand Singh
    Jul 27 '17 at 11:43






  • 16




    Results in advanced mathematics have time and again proved that one should rely more on rigor than intuition and perhaps it is time we ditch the belief that intuitive explanations are better.
    – Paramanand Singh
    Jul 27 '17 at 11:45












  • @ParamanandSingh I generally agree with you, though intuition has its place. The Greeks did Geometry with an axiomatic version of geometry based on reality that has become central to mathematics. Newton, Euler, etc. assumed the Intermediate Value Theorem and proved great results. I would even argue that familiarity to the point of intuition in a field of learning is often required to expand the field, though of course there are plenty of counterexamples and it depends on how you would define intuition. For example, I recognized immediately that the series converged...
    – Brevan Ellefsen
    Jul 27 '17 at 16:14






  • 3




    @ParamanandSingh do we associate thus with intuition or simply with prior experience? In general, is intuition unchanging or does it adapt as we become familiar with material?
    – Brevan Ellefsen
    Jul 27 '17 at 16:15










  • I like both sides of this discussion, but I think you might be using the same word to describe two different things. Intuition is basically instinct, so Paramanand is referring to the word in it's correct meaning. However, we tend to use the word intuition to refer to something that feels like instinct, but actually comes with experience. I'm not sure if there is a dedicated word for this, which is probably why we say "intuition". To a researcher in category theory, the subject is "intuitive", but to anyone else, it seems extremely esoteric and unintuitive. I think there is a way to develop...
    – AlexanderJ93
    Jul 27 '17 at 17:00















up vote
21
down vote

favorite
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$$lim_{nrightarrowinfty}biggl(frac{1}{n^2} + frac{2}{n^2} + frac{3}{n^2}+dots+frac{n}{n^2}biggr)$$




I managed to get the answer as $1$ by standard methods of solving, learned from teachers, but my intuition says that the denominator of every term grows much faster than the numerator so limit must equal to zero.



Where is my mistake? Please explain very intuitively.










share|cite|improve this question




















  • 14




    Human intuition in general is a product of evolution and history has shown that deep mathematical skills don't offer any extra advantages in human survival. Situation is similar to the fact that results of relativity and quantum mechanics are counter-intuitive. Nobody on earth had to walk with speeds close to light or deal with objects the size of an atom. Relying on intuition and trying to explain everything intuitively even at the expense of correctness is a modern trend and is based on the belief that it leads to better understanding. Cont'd..
    – Paramanand Singh
    Jul 27 '17 at 11:43






  • 16




    Results in advanced mathematics have time and again proved that one should rely more on rigor than intuition and perhaps it is time we ditch the belief that intuitive explanations are better.
    – Paramanand Singh
    Jul 27 '17 at 11:45












  • @ParamanandSingh I generally agree with you, though intuition has its place. The Greeks did Geometry with an axiomatic version of geometry based on reality that has become central to mathematics. Newton, Euler, etc. assumed the Intermediate Value Theorem and proved great results. I would even argue that familiarity to the point of intuition in a field of learning is often required to expand the field, though of course there are plenty of counterexamples and it depends on how you would define intuition. For example, I recognized immediately that the series converged...
    – Brevan Ellefsen
    Jul 27 '17 at 16:14






  • 3




    @ParamanandSingh do we associate thus with intuition or simply with prior experience? In general, is intuition unchanging or does it adapt as we become familiar with material?
    – Brevan Ellefsen
    Jul 27 '17 at 16:15










  • I like both sides of this discussion, but I think you might be using the same word to describe two different things. Intuition is basically instinct, so Paramanand is referring to the word in it's correct meaning. However, we tend to use the word intuition to refer to something that feels like instinct, but actually comes with experience. I'm not sure if there is a dedicated word for this, which is probably why we say "intuition". To a researcher in category theory, the subject is "intuitive", but to anyone else, it seems extremely esoteric and unintuitive. I think there is a way to develop...
    – AlexanderJ93
    Jul 27 '17 at 17:00













up vote
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up vote
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$$lim_{nrightarrowinfty}biggl(frac{1}{n^2} + frac{2}{n^2} + frac{3}{n^2}+dots+frac{n}{n^2}biggr)$$




I managed to get the answer as $1$ by standard methods of solving, learned from teachers, but my intuition says that the denominator of every term grows much faster than the numerator so limit must equal to zero.



Where is my mistake? Please explain very intuitively.










share|cite|improve this question
















$$lim_{nrightarrowinfty}biggl(frac{1}{n^2} + frac{2}{n^2} + frac{3}{n^2}+dots+frac{n}{n^2}biggr)$$




I managed to get the answer as $1$ by standard methods of solving, learned from teachers, but my intuition says that the denominator of every term grows much faster than the numerator so limit must equal to zero.



Where is my mistake? Please explain very intuitively.







calculus algebra-precalculus limits limits-without-lhopital






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edited Jul 28 '17 at 12:56









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asked Jul 27 '17 at 9:14









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  • 14




    Human intuition in general is a product of evolution and history has shown that deep mathematical skills don't offer any extra advantages in human survival. Situation is similar to the fact that results of relativity and quantum mechanics are counter-intuitive. Nobody on earth had to walk with speeds close to light or deal with objects the size of an atom. Relying on intuition and trying to explain everything intuitively even at the expense of correctness is a modern trend and is based on the belief that it leads to better understanding. Cont'd..
    – Paramanand Singh
    Jul 27 '17 at 11:43






  • 16




    Results in advanced mathematics have time and again proved that one should rely more on rigor than intuition and perhaps it is time we ditch the belief that intuitive explanations are better.
    – Paramanand Singh
    Jul 27 '17 at 11:45












  • @ParamanandSingh I generally agree with you, though intuition has its place. The Greeks did Geometry with an axiomatic version of geometry based on reality that has become central to mathematics. Newton, Euler, etc. assumed the Intermediate Value Theorem and proved great results. I would even argue that familiarity to the point of intuition in a field of learning is often required to expand the field, though of course there are plenty of counterexamples and it depends on how you would define intuition. For example, I recognized immediately that the series converged...
    – Brevan Ellefsen
    Jul 27 '17 at 16:14






  • 3




    @ParamanandSingh do we associate thus with intuition or simply with prior experience? In general, is intuition unchanging or does it adapt as we become familiar with material?
    – Brevan Ellefsen
    Jul 27 '17 at 16:15










  • I like both sides of this discussion, but I think you might be using the same word to describe two different things. Intuition is basically instinct, so Paramanand is referring to the word in it's correct meaning. However, we tend to use the word intuition to refer to something that feels like instinct, but actually comes with experience. I'm not sure if there is a dedicated word for this, which is probably why we say "intuition". To a researcher in category theory, the subject is "intuitive", but to anyone else, it seems extremely esoteric and unintuitive. I think there is a way to develop...
    – AlexanderJ93
    Jul 27 '17 at 17:00














  • 14




    Human intuition in general is a product of evolution and history has shown that deep mathematical skills don't offer any extra advantages in human survival. Situation is similar to the fact that results of relativity and quantum mechanics are counter-intuitive. Nobody on earth had to walk with speeds close to light or deal with objects the size of an atom. Relying on intuition and trying to explain everything intuitively even at the expense of correctness is a modern trend and is based on the belief that it leads to better understanding. Cont'd..
    – Paramanand Singh
    Jul 27 '17 at 11:43






  • 16




    Results in advanced mathematics have time and again proved that one should rely more on rigor than intuition and perhaps it is time we ditch the belief that intuitive explanations are better.
    – Paramanand Singh
    Jul 27 '17 at 11:45












  • @ParamanandSingh I generally agree with you, though intuition has its place. The Greeks did Geometry with an axiomatic version of geometry based on reality that has become central to mathematics. Newton, Euler, etc. assumed the Intermediate Value Theorem and proved great results. I would even argue that familiarity to the point of intuition in a field of learning is often required to expand the field, though of course there are plenty of counterexamples and it depends on how you would define intuition. For example, I recognized immediately that the series converged...
    – Brevan Ellefsen
    Jul 27 '17 at 16:14






  • 3




    @ParamanandSingh do we associate thus with intuition or simply with prior experience? In general, is intuition unchanging or does it adapt as we become familiar with material?
    – Brevan Ellefsen
    Jul 27 '17 at 16:15










  • I like both sides of this discussion, but I think you might be using the same word to describe two different things. Intuition is basically instinct, so Paramanand is referring to the word in it's correct meaning. However, we tend to use the word intuition to refer to something that feels like instinct, but actually comes with experience. I'm not sure if there is a dedicated word for this, which is probably why we say "intuition". To a researcher in category theory, the subject is "intuitive", but to anyone else, it seems extremely esoteric and unintuitive. I think there is a way to develop...
    – AlexanderJ93
    Jul 27 '17 at 17:00








14




14




Human intuition in general is a product of evolution and history has shown that deep mathematical skills don't offer any extra advantages in human survival. Situation is similar to the fact that results of relativity and quantum mechanics are counter-intuitive. Nobody on earth had to walk with speeds close to light or deal with objects the size of an atom. Relying on intuition and trying to explain everything intuitively even at the expense of correctness is a modern trend and is based on the belief that it leads to better understanding. Cont'd..
– Paramanand Singh
Jul 27 '17 at 11:43




Human intuition in general is a product of evolution and history has shown that deep mathematical skills don't offer any extra advantages in human survival. Situation is similar to the fact that results of relativity and quantum mechanics are counter-intuitive. Nobody on earth had to walk with speeds close to light or deal with objects the size of an atom. Relying on intuition and trying to explain everything intuitively even at the expense of correctness is a modern trend and is based on the belief that it leads to better understanding. Cont'd..
– Paramanand Singh
Jul 27 '17 at 11:43




16




16




Results in advanced mathematics have time and again proved that one should rely more on rigor than intuition and perhaps it is time we ditch the belief that intuitive explanations are better.
– Paramanand Singh
Jul 27 '17 at 11:45






Results in advanced mathematics have time and again proved that one should rely more on rigor than intuition and perhaps it is time we ditch the belief that intuitive explanations are better.
– Paramanand Singh
Jul 27 '17 at 11:45














@ParamanandSingh I generally agree with you, though intuition has its place. The Greeks did Geometry with an axiomatic version of geometry based on reality that has become central to mathematics. Newton, Euler, etc. assumed the Intermediate Value Theorem and proved great results. I would even argue that familiarity to the point of intuition in a field of learning is often required to expand the field, though of course there are plenty of counterexamples and it depends on how you would define intuition. For example, I recognized immediately that the series converged...
– Brevan Ellefsen
Jul 27 '17 at 16:14




@ParamanandSingh I generally agree with you, though intuition has its place. The Greeks did Geometry with an axiomatic version of geometry based on reality that has become central to mathematics. Newton, Euler, etc. assumed the Intermediate Value Theorem and proved great results. I would even argue that familiarity to the point of intuition in a field of learning is often required to expand the field, though of course there are plenty of counterexamples and it depends on how you would define intuition. For example, I recognized immediately that the series converged...
– Brevan Ellefsen
Jul 27 '17 at 16:14




3




3




@ParamanandSingh do we associate thus with intuition or simply with prior experience? In general, is intuition unchanging or does it adapt as we become familiar with material?
– Brevan Ellefsen
Jul 27 '17 at 16:15




@ParamanandSingh do we associate thus with intuition or simply with prior experience? In general, is intuition unchanging or does it adapt as we become familiar with material?
– Brevan Ellefsen
Jul 27 '17 at 16:15












I like both sides of this discussion, but I think you might be using the same word to describe two different things. Intuition is basically instinct, so Paramanand is referring to the word in it's correct meaning. However, we tend to use the word intuition to refer to something that feels like instinct, but actually comes with experience. I'm not sure if there is a dedicated word for this, which is probably why we say "intuition". To a researcher in category theory, the subject is "intuitive", but to anyone else, it seems extremely esoteric and unintuitive. I think there is a way to develop...
– AlexanderJ93
Jul 27 '17 at 17:00




I like both sides of this discussion, but I think you might be using the same word to describe two different things. Intuition is basically instinct, so Paramanand is referring to the word in it's correct meaning. However, we tend to use the word intuition to refer to something that feels like instinct, but actually comes with experience. I'm not sure if there is a dedicated word for this, which is probably why we say "intuition". To a researcher in category theory, the subject is "intuitive", but to anyone else, it seems extremely esoteric and unintuitive. I think there is a way to develop...
– AlexanderJ93
Jul 27 '17 at 17:00










11 Answers
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Intuition should say:



the denominator grows with $n^2$, the numerator grows with $n$. However, the number of fractions also grows by $n$, so the total growth of the numerator is about $n^2$.





And that's where intuition stops. From here on, you go with logic and rigor, not intuition.



And it gets you to



$$frac1{n^2} + frac2{n^2}+cdots + frac{n}{n^2} = frac{1+2+3+cdots + n}{n^2} = frac{frac{n(n+1)}{2}}{n^2} = frac{n^2+n}{2n^2}$$



and you find that the limit is $frac12$ (not $1$!)






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    up vote
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    With integrals:
    $lim_{nrightarrowinfty}biggl(frac{1}{n^2} + frac{2}{n^2} + frac{3}{n^2}+dots+frac{n}{n^2}biggr)=lim_{nrightarrowinfty} frac{1}{n}biggl(frac{1}{n} + frac{2}{n} + frac{3}{n}+dots+frac{n}{n}biggr)= int_{0}^1 x dx =frac{1}{2}$.






    share|cite|improve this answer




























      up vote
      15
      down vote













      Notice that:



      $$lim_{nrightarrowinfty}dfrac{1}{n^2} + dfrac{2}{n^2} + dfrac{3}{n^2}+cdots+dfrac{n}{n^2}=lim_{nrightarrowinfty} frac{1}{n^2}left(1+2+dots nright)$$
      and this last sum can be replaced by the Gauss formula, so it becomes:
      $$lim_{nrightarrowinfty}frac{1}{n^2}left(frac{n(n+1)}{2}right)$$



      $$=lim_{nrightarrowinfty}frac{n+1}{2n}=frac{1}{2}left(lim_{nrightarrowinfty} frac{n+1}{n}right)=frac{1}{2}left(lim_{nrightarrowinfty} 1+frac{1}{n}right)=frac{1}{2}$$






      share|cite|improve this answer






























        up vote
        10
        down vote













        Perhaps this is beyond your curriculum, but one could also use the Stolz-Cesàro theorem.



        $$lim_{ntoinfty}frac{1+2+cdots+(n-1)+n}{n^2}$$



        Denote the numerator and denominator, respectively, by



        $$a_n=sum_{k=1}^nk$$
        $$b_n=n^2$$



        It's easy to show that $b_n$ is strictly monotone and divergent. Stolz-Cesàro then says that if



        $$lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$



        exists, then it is equal to



        $$lim_{ntoinfty}frac{a_n}{b_n}$$



        We have



        $$lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}=lim_{ntoinfty}frac{sumlimits_{k=1}^{n+1}k-sumlimits_{k=1}^nk}{(n+1)^2-n^2}=lim_{ntoinfty}frac{n+1}{2n+1}=frac12$$





        As the Wiki page mentions, the theorem is sometimes referred to as "L'Hopital's rule for sequences". If you find that intuitive (related question), then perhaps you will think the same of the approach using this theorem.






        share|cite|improve this answer






























          up vote
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          down vote













          If you pull out the denominator, the numerator sums to $frac{n(n+1)}{2}$, then $n^2$ term cancels out and when you take the limit you get $frac{1}{2}$






          share|cite|improve this answer




























            up vote
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            Your intuition "the denominator of every term grows much faster than the numerator" is partially true. It would be completely true if there were finite number of terms. However, you must bear in mind that there are infinite number of terms. Hence you have two variables: denominator $n^2$ and numerator $1+2+cdots+n$ (when you add up the numerators of the fractions with common denominator). Dealing with infinitely small or large quantities is frequently risky and counterintuitive, because in the given limit you have so called indetermined form of $frac{infty}{infty}$ or $0cdot infty$ (if expressed as $frac{1}{n^2}cdot (1+2+cdots+n)$). Thus the result of such indefinite forms depends on their relative rates of increase and/or decrease. As an example, consider a rectangle of area $A$. If its width is reduced infinitely while its length is held constant, then its area will decrease to $0$. If its width is reduced while its length is extended at the same rate, then its area will stay constant. This is what happenned to your limit. And if its width is reduced while its length is extended at a relatively higher rate, the its area will increase infinitely.



            Keep developing your intuition!






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              up vote
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              You have asked where you are mistaken, so I will only answer this. You have tried to apply the limit individually to the summands. There is a rule which allows you to do this for finite sums, but here we do not have a fixed number of summands, and this rule does not apply. For a simpler example why you also cannot expect it to apply, consider
              $$
              lim_{ntoinfty}Bigl(underbrace{frac1n+frac1n+dots+frac1n}_{text{$n$ summands}}Bigr).
              $$
              You could again reason that each summand goes to zero, and therefore so must the sum. But of course this is just a fancy way of writing
              $$lim_{ntoinfty} 1,$$
              and the limit is $1$. So you see that in general you cannot apply the limits individually to the summands of a sum if the number of summands is not fixed and finite.





              Indeed by adding zeros on the right you can view the sum as a series. A series is just a limit of partial sums, and so this also gives you a nice example that you cannot in general exchange two limits.






              share|cite|improve this answer




























                up vote
                3
                down vote













                Seeing such a term, my intuition would be:




                • $n²$ in the denominator, yup, grows quickly (although, not that quickly either, compared to $x^n$ or something like that).

                • But, danger!, there are also growing numbers of terms in the nominator.


                The second line is the important one; this immediately tells me that I cannot trust any further intuition (i.e., random ideas shooting spontaneously through my mind) anymore, and that it is time to find a pen and some paper.



                Full disclosure: I am more at home in CS than maths, but the same is true for O notation in CS, it can be quite misleading and unintuitive. Or rather, deceivingly simple and provoking intuitive errors.






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                  up vote
                  3
                  down vote













                  "grows much faster than the numerator": you are disregarding the fact that the numerator is actually



                  $$1+2+3+cdots n, $$ which doesn't grow slower than $n^2$.






                  share|cite|improve this answer




























                    up vote
                    0
                    down vote













                    Very intuitively, there are a lot of positive terms inside the brackets, so they can add up to something non-zero. An extremely loose way to get at what is going on is to over estimate by $n$ times the largest term and under estimate $n/2$ times the middle term, so you have



                    $$
                    frac{1}{4} = frac{n}{2}cdot frac{n/2}{n^2} le frac{1}{n^2} + frac{2}{n^2} + cdots + frac{n-1}{n^2} + frac{n}{n^2} le ncdot frac{n}{n^2} = 1
                    $$



                    In fact, the answer is just the limit of $n^{-2}binom{n+1}{2} = frac{1}{2}$ as $ntoinfty$, but the above is why you should guess that the answer isn't zero.






                    share|cite|improve this answer




























                      up vote
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                      down vote













                      Here's an explanation based on the intuitively obvious fact that a triangle with very little height can nonetheless have substantial area provided it has a very long base.



                      Draw the step function $f_n(x)={lceil xrceilover n^2}$ for $0le xle n$, and observe that it lies above the triangle with vertices at $(0,0)$, $(n,0)$ and $(n,{1over n})$ -- i.e., a right triangle with base of length $n$ and height $1over n$, hence area $1over2$. Since the area beneath the step function is $S_n={1over n^2}+{2over n^2}+cdots+{nover n^2}$, this shows that $lim_{ntoinfty}S_n$, if it exists, cannot be less than $1over2$.



                      If you like, you can also draw the step function $g_n(x)={lfloor xrfloorover n^2}$ for $0le xle n+1$ which lies below a right triangle with base of length $n+1$ and height $n+1over n^2$ and argue that the area beneath the step function, ${0over n^2}+{1over n^2}+{2over n^2}+cdots+{nover n^2}$, is less than ${1over2}left(n+1over nright)^2$. Combining this with the result of the first paragraph leads to the limit $S_nto{1over2}$.






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                        11 Answers
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                        11 Answers
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                        active

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                        up vote
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                        down vote



                        accepted










                        Intuition should say:



                        the denominator grows with $n^2$, the numerator grows with $n$. However, the number of fractions also grows by $n$, so the total growth of the numerator is about $n^2$.





                        And that's where intuition stops. From here on, you go with logic and rigor, not intuition.



                        And it gets you to



                        $$frac1{n^2} + frac2{n^2}+cdots + frac{n}{n^2} = frac{1+2+3+cdots + n}{n^2} = frac{frac{n(n+1)}{2}}{n^2} = frac{n^2+n}{2n^2}$$



                        and you find that the limit is $frac12$ (not $1$!)






                        share|cite|improve this answer

























                          up vote
                          74
                          down vote



                          accepted










                          Intuition should say:



                          the denominator grows with $n^2$, the numerator grows with $n$. However, the number of fractions also grows by $n$, so the total growth of the numerator is about $n^2$.





                          And that's where intuition stops. From here on, you go with logic and rigor, not intuition.



                          And it gets you to



                          $$frac1{n^2} + frac2{n^2}+cdots + frac{n}{n^2} = frac{1+2+3+cdots + n}{n^2} = frac{frac{n(n+1)}{2}}{n^2} = frac{n^2+n}{2n^2}$$



                          and you find that the limit is $frac12$ (not $1$!)






                          share|cite|improve this answer























                            up vote
                            74
                            down vote



                            accepted







                            up vote
                            74
                            down vote



                            accepted






                            Intuition should say:



                            the denominator grows with $n^2$, the numerator grows with $n$. However, the number of fractions also grows by $n$, so the total growth of the numerator is about $n^2$.





                            And that's where intuition stops. From here on, you go with logic and rigor, not intuition.



                            And it gets you to



                            $$frac1{n^2} + frac2{n^2}+cdots + frac{n}{n^2} = frac{1+2+3+cdots + n}{n^2} = frac{frac{n(n+1)}{2}}{n^2} = frac{n^2+n}{2n^2}$$



                            and you find that the limit is $frac12$ (not $1$!)






                            share|cite|improve this answer












                            Intuition should say:



                            the denominator grows with $n^2$, the numerator grows with $n$. However, the number of fractions also grows by $n$, so the total growth of the numerator is about $n^2$.





                            And that's where intuition stops. From here on, you go with logic and rigor, not intuition.



                            And it gets you to



                            $$frac1{n^2} + frac2{n^2}+cdots + frac{n}{n^2} = frac{1+2+3+cdots + n}{n^2} = frac{frac{n(n+1)}{2}}{n^2} = frac{n^2+n}{2n^2}$$



                            and you find that the limit is $frac12$ (not $1$!)







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jul 27 '17 at 9:29









                            5xum

                            89.4k393161




                            89.4k393161






















                                up vote
                                33
                                down vote













                                With integrals:
                                $lim_{nrightarrowinfty}biggl(frac{1}{n^2} + frac{2}{n^2} + frac{3}{n^2}+dots+frac{n}{n^2}biggr)=lim_{nrightarrowinfty} frac{1}{n}biggl(frac{1}{n} + frac{2}{n} + frac{3}{n}+dots+frac{n}{n}biggr)= int_{0}^1 x dx =frac{1}{2}$.






                                share|cite|improve this answer

























                                  up vote
                                  33
                                  down vote













                                  With integrals:
                                  $lim_{nrightarrowinfty}biggl(frac{1}{n^2} + frac{2}{n^2} + frac{3}{n^2}+dots+frac{n}{n^2}biggr)=lim_{nrightarrowinfty} frac{1}{n}biggl(frac{1}{n} + frac{2}{n} + frac{3}{n}+dots+frac{n}{n}biggr)= int_{0}^1 x dx =frac{1}{2}$.






                                  share|cite|improve this answer























                                    up vote
                                    33
                                    down vote










                                    up vote
                                    33
                                    down vote









                                    With integrals:
                                    $lim_{nrightarrowinfty}biggl(frac{1}{n^2} + frac{2}{n^2} + frac{3}{n^2}+dots+frac{n}{n^2}biggr)=lim_{nrightarrowinfty} frac{1}{n}biggl(frac{1}{n} + frac{2}{n} + frac{3}{n}+dots+frac{n}{n}biggr)= int_{0}^1 x dx =frac{1}{2}$.






                                    share|cite|improve this answer












                                    With integrals:
                                    $lim_{nrightarrowinfty}biggl(frac{1}{n^2} + frac{2}{n^2} + frac{3}{n^2}+dots+frac{n}{n^2}biggr)=lim_{nrightarrowinfty} frac{1}{n}biggl(frac{1}{n} + frac{2}{n} + frac{3}{n}+dots+frac{n}{n}biggr)= int_{0}^1 x dx =frac{1}{2}$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jul 27 '17 at 10:24









                                    Fred

                                    43.6k1644




                                    43.6k1644






















                                        up vote
                                        15
                                        down vote













                                        Notice that:



                                        $$lim_{nrightarrowinfty}dfrac{1}{n^2} + dfrac{2}{n^2} + dfrac{3}{n^2}+cdots+dfrac{n}{n^2}=lim_{nrightarrowinfty} frac{1}{n^2}left(1+2+dots nright)$$
                                        and this last sum can be replaced by the Gauss formula, so it becomes:
                                        $$lim_{nrightarrowinfty}frac{1}{n^2}left(frac{n(n+1)}{2}right)$$



                                        $$=lim_{nrightarrowinfty}frac{n+1}{2n}=frac{1}{2}left(lim_{nrightarrowinfty} frac{n+1}{n}right)=frac{1}{2}left(lim_{nrightarrowinfty} 1+frac{1}{n}right)=frac{1}{2}$$






                                        share|cite|improve this answer



























                                          up vote
                                          15
                                          down vote













                                          Notice that:



                                          $$lim_{nrightarrowinfty}dfrac{1}{n^2} + dfrac{2}{n^2} + dfrac{3}{n^2}+cdots+dfrac{n}{n^2}=lim_{nrightarrowinfty} frac{1}{n^2}left(1+2+dots nright)$$
                                          and this last sum can be replaced by the Gauss formula, so it becomes:
                                          $$lim_{nrightarrowinfty}frac{1}{n^2}left(frac{n(n+1)}{2}right)$$



                                          $$=lim_{nrightarrowinfty}frac{n+1}{2n}=frac{1}{2}left(lim_{nrightarrowinfty} frac{n+1}{n}right)=frac{1}{2}left(lim_{nrightarrowinfty} 1+frac{1}{n}right)=frac{1}{2}$$






                                          share|cite|improve this answer

























                                            up vote
                                            15
                                            down vote










                                            up vote
                                            15
                                            down vote









                                            Notice that:



                                            $$lim_{nrightarrowinfty}dfrac{1}{n^2} + dfrac{2}{n^2} + dfrac{3}{n^2}+cdots+dfrac{n}{n^2}=lim_{nrightarrowinfty} frac{1}{n^2}left(1+2+dots nright)$$
                                            and this last sum can be replaced by the Gauss formula, so it becomes:
                                            $$lim_{nrightarrowinfty}frac{1}{n^2}left(frac{n(n+1)}{2}right)$$



                                            $$=lim_{nrightarrowinfty}frac{n+1}{2n}=frac{1}{2}left(lim_{nrightarrowinfty} frac{n+1}{n}right)=frac{1}{2}left(lim_{nrightarrowinfty} 1+frac{1}{n}right)=frac{1}{2}$$






                                            share|cite|improve this answer














                                            Notice that:



                                            $$lim_{nrightarrowinfty}dfrac{1}{n^2} + dfrac{2}{n^2} + dfrac{3}{n^2}+cdots+dfrac{n}{n^2}=lim_{nrightarrowinfty} frac{1}{n^2}left(1+2+dots nright)$$
                                            and this last sum can be replaced by the Gauss formula, so it becomes:
                                            $$lim_{nrightarrowinfty}frac{1}{n^2}left(frac{n(n+1)}{2}right)$$



                                            $$=lim_{nrightarrowinfty}frac{n+1}{2n}=frac{1}{2}left(lim_{nrightarrowinfty} frac{n+1}{n}right)=frac{1}{2}left(lim_{nrightarrowinfty} 1+frac{1}{n}right)=frac{1}{2}$$







                                            share|cite|improve this answer














                                            share|cite|improve this answer



                                            share|cite|improve this answer








                                            edited Nov 27 at 22:48

























                                            answered Jul 27 '17 at 9:17









                                            MonsieurGalois

                                            3,3981333




                                            3,3981333






















                                                up vote
                                                10
                                                down vote













                                                Perhaps this is beyond your curriculum, but one could also use the Stolz-Cesàro theorem.



                                                $$lim_{ntoinfty}frac{1+2+cdots+(n-1)+n}{n^2}$$



                                                Denote the numerator and denominator, respectively, by



                                                $$a_n=sum_{k=1}^nk$$
                                                $$b_n=n^2$$



                                                It's easy to show that $b_n$ is strictly monotone and divergent. Stolz-Cesàro then says that if



                                                $$lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$



                                                exists, then it is equal to



                                                $$lim_{ntoinfty}frac{a_n}{b_n}$$



                                                We have



                                                $$lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}=lim_{ntoinfty}frac{sumlimits_{k=1}^{n+1}k-sumlimits_{k=1}^nk}{(n+1)^2-n^2}=lim_{ntoinfty}frac{n+1}{2n+1}=frac12$$





                                                As the Wiki page mentions, the theorem is sometimes referred to as "L'Hopital's rule for sequences". If you find that intuitive (related question), then perhaps you will think the same of the approach using this theorem.






                                                share|cite|improve this answer



























                                                  up vote
                                                  10
                                                  down vote













                                                  Perhaps this is beyond your curriculum, but one could also use the Stolz-Cesàro theorem.



                                                  $$lim_{ntoinfty}frac{1+2+cdots+(n-1)+n}{n^2}$$



                                                  Denote the numerator and denominator, respectively, by



                                                  $$a_n=sum_{k=1}^nk$$
                                                  $$b_n=n^2$$



                                                  It's easy to show that $b_n$ is strictly monotone and divergent. Stolz-Cesàro then says that if



                                                  $$lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$



                                                  exists, then it is equal to



                                                  $$lim_{ntoinfty}frac{a_n}{b_n}$$



                                                  We have



                                                  $$lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}=lim_{ntoinfty}frac{sumlimits_{k=1}^{n+1}k-sumlimits_{k=1}^nk}{(n+1)^2-n^2}=lim_{ntoinfty}frac{n+1}{2n+1}=frac12$$





                                                  As the Wiki page mentions, the theorem is sometimes referred to as "L'Hopital's rule for sequences". If you find that intuitive (related question), then perhaps you will think the same of the approach using this theorem.






                                                  share|cite|improve this answer

























                                                    up vote
                                                    10
                                                    down vote










                                                    up vote
                                                    10
                                                    down vote









                                                    Perhaps this is beyond your curriculum, but one could also use the Stolz-Cesàro theorem.



                                                    $$lim_{ntoinfty}frac{1+2+cdots+(n-1)+n}{n^2}$$



                                                    Denote the numerator and denominator, respectively, by



                                                    $$a_n=sum_{k=1}^nk$$
                                                    $$b_n=n^2$$



                                                    It's easy to show that $b_n$ is strictly monotone and divergent. Stolz-Cesàro then says that if



                                                    $$lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$



                                                    exists, then it is equal to



                                                    $$lim_{ntoinfty}frac{a_n}{b_n}$$



                                                    We have



                                                    $$lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}=lim_{ntoinfty}frac{sumlimits_{k=1}^{n+1}k-sumlimits_{k=1}^nk}{(n+1)^2-n^2}=lim_{ntoinfty}frac{n+1}{2n+1}=frac12$$





                                                    As the Wiki page mentions, the theorem is sometimes referred to as "L'Hopital's rule for sequences". If you find that intuitive (related question), then perhaps you will think the same of the approach using this theorem.






                                                    share|cite|improve this answer














                                                    Perhaps this is beyond your curriculum, but one could also use the Stolz-Cesàro theorem.



                                                    $$lim_{ntoinfty}frac{1+2+cdots+(n-1)+n}{n^2}$$



                                                    Denote the numerator and denominator, respectively, by



                                                    $$a_n=sum_{k=1}^nk$$
                                                    $$b_n=n^2$$



                                                    It's easy to show that $b_n$ is strictly monotone and divergent. Stolz-Cesàro then says that if



                                                    $$lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$



                                                    exists, then it is equal to



                                                    $$lim_{ntoinfty}frac{a_n}{b_n}$$



                                                    We have



                                                    $$lim_{ntoinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}=lim_{ntoinfty}frac{sumlimits_{k=1}^{n+1}k-sumlimits_{k=1}^nk}{(n+1)^2-n^2}=lim_{ntoinfty}frac{n+1}{2n+1}=frac12$$





                                                    As the Wiki page mentions, the theorem is sometimes referred to as "L'Hopital's rule for sequences". If you find that intuitive (related question), then perhaps you will think the same of the approach using this theorem.







                                                    share|cite|improve this answer














                                                    share|cite|improve this answer



                                                    share|cite|improve this answer








                                                    edited Sep 29 '17 at 19:47

























                                                    answered Jul 27 '17 at 16:49









                                                    user170231

                                                    3,96711229




                                                    3,96711229






















                                                        up vote
                                                        9
                                                        down vote













                                                        If you pull out the denominator, the numerator sums to $frac{n(n+1)}{2}$, then $n^2$ term cancels out and when you take the limit you get $frac{1}{2}$






                                                        share|cite|improve this answer

























                                                          up vote
                                                          9
                                                          down vote













                                                          If you pull out the denominator, the numerator sums to $frac{n(n+1)}{2}$, then $n^2$ term cancels out and when you take the limit you get $frac{1}{2}$






                                                          share|cite|improve this answer























                                                            up vote
                                                            9
                                                            down vote










                                                            up vote
                                                            9
                                                            down vote









                                                            If you pull out the denominator, the numerator sums to $frac{n(n+1)}{2}$, then $n^2$ term cancels out and when you take the limit you get $frac{1}{2}$






                                                            share|cite|improve this answer












                                                            If you pull out the denominator, the numerator sums to $frac{n(n+1)}{2}$, then $n^2$ term cancels out and when you take the limit you get $frac{1}{2}$







                                                            share|cite|improve this answer












                                                            share|cite|improve this answer



                                                            share|cite|improve this answer










                                                            answered Jul 27 '17 at 9:17









                                                            Alex

                                                            14.2k42134




                                                            14.2k42134






















                                                                up vote
                                                                4
                                                                down vote













                                                                Your intuition "the denominator of every term grows much faster than the numerator" is partially true. It would be completely true if there were finite number of terms. However, you must bear in mind that there are infinite number of terms. Hence you have two variables: denominator $n^2$ and numerator $1+2+cdots+n$ (when you add up the numerators of the fractions with common denominator). Dealing with infinitely small or large quantities is frequently risky and counterintuitive, because in the given limit you have so called indetermined form of $frac{infty}{infty}$ or $0cdot infty$ (if expressed as $frac{1}{n^2}cdot (1+2+cdots+n)$). Thus the result of such indefinite forms depends on their relative rates of increase and/or decrease. As an example, consider a rectangle of area $A$. If its width is reduced infinitely while its length is held constant, then its area will decrease to $0$. If its width is reduced while its length is extended at the same rate, then its area will stay constant. This is what happenned to your limit. And if its width is reduced while its length is extended at a relatively higher rate, the its area will increase infinitely.



                                                                Keep developing your intuition!






                                                                share|cite|improve this answer

























                                                                  up vote
                                                                  4
                                                                  down vote













                                                                  Your intuition "the denominator of every term grows much faster than the numerator" is partially true. It would be completely true if there were finite number of terms. However, you must bear in mind that there are infinite number of terms. Hence you have two variables: denominator $n^2$ and numerator $1+2+cdots+n$ (when you add up the numerators of the fractions with common denominator). Dealing with infinitely small or large quantities is frequently risky and counterintuitive, because in the given limit you have so called indetermined form of $frac{infty}{infty}$ or $0cdot infty$ (if expressed as $frac{1}{n^2}cdot (1+2+cdots+n)$). Thus the result of such indefinite forms depends on their relative rates of increase and/or decrease. As an example, consider a rectangle of area $A$. If its width is reduced infinitely while its length is held constant, then its area will decrease to $0$. If its width is reduced while its length is extended at the same rate, then its area will stay constant. This is what happenned to your limit. And if its width is reduced while its length is extended at a relatively higher rate, the its area will increase infinitely.



                                                                  Keep developing your intuition!






                                                                  share|cite|improve this answer























                                                                    up vote
                                                                    4
                                                                    down vote










                                                                    up vote
                                                                    4
                                                                    down vote









                                                                    Your intuition "the denominator of every term grows much faster than the numerator" is partially true. It would be completely true if there were finite number of terms. However, you must bear in mind that there are infinite number of terms. Hence you have two variables: denominator $n^2$ and numerator $1+2+cdots+n$ (when you add up the numerators of the fractions with common denominator). Dealing with infinitely small or large quantities is frequently risky and counterintuitive, because in the given limit you have so called indetermined form of $frac{infty}{infty}$ or $0cdot infty$ (if expressed as $frac{1}{n^2}cdot (1+2+cdots+n)$). Thus the result of such indefinite forms depends on their relative rates of increase and/or decrease. As an example, consider a rectangle of area $A$. If its width is reduced infinitely while its length is held constant, then its area will decrease to $0$. If its width is reduced while its length is extended at the same rate, then its area will stay constant. This is what happenned to your limit. And if its width is reduced while its length is extended at a relatively higher rate, the its area will increase infinitely.



                                                                    Keep developing your intuition!






                                                                    share|cite|improve this answer












                                                                    Your intuition "the denominator of every term grows much faster than the numerator" is partially true. It would be completely true if there were finite number of terms. However, you must bear in mind that there are infinite number of terms. Hence you have two variables: denominator $n^2$ and numerator $1+2+cdots+n$ (when you add up the numerators of the fractions with common denominator). Dealing with infinitely small or large quantities is frequently risky and counterintuitive, because in the given limit you have so called indetermined form of $frac{infty}{infty}$ or $0cdot infty$ (if expressed as $frac{1}{n^2}cdot (1+2+cdots+n)$). Thus the result of such indefinite forms depends on their relative rates of increase and/or decrease. As an example, consider a rectangle of area $A$. If its width is reduced infinitely while its length is held constant, then its area will decrease to $0$. If its width is reduced while its length is extended at the same rate, then its area will stay constant. This is what happenned to your limit. And if its width is reduced while its length is extended at a relatively higher rate, the its area will increase infinitely.



                                                                    Keep developing your intuition!







                                                                    share|cite|improve this answer












                                                                    share|cite|improve this answer



                                                                    share|cite|improve this answer










                                                                    answered Jul 27 '17 at 20:28









                                                                    farruhota

                                                                    18.7k2736




                                                                    18.7k2736






















                                                                        up vote
                                                                        4
                                                                        down vote













                                                                        You have asked where you are mistaken, so I will only answer this. You have tried to apply the limit individually to the summands. There is a rule which allows you to do this for finite sums, but here we do not have a fixed number of summands, and this rule does not apply. For a simpler example why you also cannot expect it to apply, consider
                                                                        $$
                                                                        lim_{ntoinfty}Bigl(underbrace{frac1n+frac1n+dots+frac1n}_{text{$n$ summands}}Bigr).
                                                                        $$
                                                                        You could again reason that each summand goes to zero, and therefore so must the sum. But of course this is just a fancy way of writing
                                                                        $$lim_{ntoinfty} 1,$$
                                                                        and the limit is $1$. So you see that in general you cannot apply the limits individually to the summands of a sum if the number of summands is not fixed and finite.





                                                                        Indeed by adding zeros on the right you can view the sum as a series. A series is just a limit of partial sums, and so this also gives you a nice example that you cannot in general exchange two limits.






                                                                        share|cite|improve this answer

























                                                                          up vote
                                                                          4
                                                                          down vote













                                                                          You have asked where you are mistaken, so I will only answer this. You have tried to apply the limit individually to the summands. There is a rule which allows you to do this for finite sums, but here we do not have a fixed number of summands, and this rule does not apply. For a simpler example why you also cannot expect it to apply, consider
                                                                          $$
                                                                          lim_{ntoinfty}Bigl(underbrace{frac1n+frac1n+dots+frac1n}_{text{$n$ summands}}Bigr).
                                                                          $$
                                                                          You could again reason that each summand goes to zero, and therefore so must the sum. But of course this is just a fancy way of writing
                                                                          $$lim_{ntoinfty} 1,$$
                                                                          and the limit is $1$. So you see that in general you cannot apply the limits individually to the summands of a sum if the number of summands is not fixed and finite.





                                                                          Indeed by adding zeros on the right you can view the sum as a series. A series is just a limit of partial sums, and so this also gives you a nice example that you cannot in general exchange two limits.






                                                                          share|cite|improve this answer























                                                                            up vote
                                                                            4
                                                                            down vote










                                                                            up vote
                                                                            4
                                                                            down vote









                                                                            You have asked where you are mistaken, so I will only answer this. You have tried to apply the limit individually to the summands. There is a rule which allows you to do this for finite sums, but here we do not have a fixed number of summands, and this rule does not apply. For a simpler example why you also cannot expect it to apply, consider
                                                                            $$
                                                                            lim_{ntoinfty}Bigl(underbrace{frac1n+frac1n+dots+frac1n}_{text{$n$ summands}}Bigr).
                                                                            $$
                                                                            You could again reason that each summand goes to zero, and therefore so must the sum. But of course this is just a fancy way of writing
                                                                            $$lim_{ntoinfty} 1,$$
                                                                            and the limit is $1$. So you see that in general you cannot apply the limits individually to the summands of a sum if the number of summands is not fixed and finite.





                                                                            Indeed by adding zeros on the right you can view the sum as a series. A series is just a limit of partial sums, and so this also gives you a nice example that you cannot in general exchange two limits.






                                                                            share|cite|improve this answer












                                                                            You have asked where you are mistaken, so I will only answer this. You have tried to apply the limit individually to the summands. There is a rule which allows you to do this for finite sums, but here we do not have a fixed number of summands, and this rule does not apply. For a simpler example why you also cannot expect it to apply, consider
                                                                            $$
                                                                            lim_{ntoinfty}Bigl(underbrace{frac1n+frac1n+dots+frac1n}_{text{$n$ summands}}Bigr).
                                                                            $$
                                                                            You could again reason that each summand goes to zero, and therefore so must the sum. But of course this is just a fancy way of writing
                                                                            $$lim_{ntoinfty} 1,$$
                                                                            and the limit is $1$. So you see that in general you cannot apply the limits individually to the summands of a sum if the number of summands is not fixed and finite.





                                                                            Indeed by adding zeros on the right you can view the sum as a series. A series is just a limit of partial sums, and so this also gives you a nice example that you cannot in general exchange two limits.







                                                                            share|cite|improve this answer












                                                                            share|cite|improve this answer



                                                                            share|cite|improve this answer










                                                                            answered Jul 28 '17 at 8:49









                                                                            Carsten S

                                                                            6,84311334




                                                                            6,84311334






















                                                                                up vote
                                                                                3
                                                                                down vote













                                                                                Seeing such a term, my intuition would be:




                                                                                • $n²$ in the denominator, yup, grows quickly (although, not that quickly either, compared to $x^n$ or something like that).

                                                                                • But, danger!, there are also growing numbers of terms in the nominator.


                                                                                The second line is the important one; this immediately tells me that I cannot trust any further intuition (i.e., random ideas shooting spontaneously through my mind) anymore, and that it is time to find a pen and some paper.



                                                                                Full disclosure: I am more at home in CS than maths, but the same is true for O notation in CS, it can be quite misleading and unintuitive. Or rather, deceivingly simple and provoking intuitive errors.






                                                                                share|cite|improve this answer

























                                                                                  up vote
                                                                                  3
                                                                                  down vote













                                                                                  Seeing such a term, my intuition would be:




                                                                                  • $n²$ in the denominator, yup, grows quickly (although, not that quickly either, compared to $x^n$ or something like that).

                                                                                  • But, danger!, there are also growing numbers of terms in the nominator.


                                                                                  The second line is the important one; this immediately tells me that I cannot trust any further intuition (i.e., random ideas shooting spontaneously through my mind) anymore, and that it is time to find a pen and some paper.



                                                                                  Full disclosure: I am more at home in CS than maths, but the same is true for O notation in CS, it can be quite misleading and unintuitive. Or rather, deceivingly simple and provoking intuitive errors.






                                                                                  share|cite|improve this answer























                                                                                    up vote
                                                                                    3
                                                                                    down vote










                                                                                    up vote
                                                                                    3
                                                                                    down vote









                                                                                    Seeing such a term, my intuition would be:




                                                                                    • $n²$ in the denominator, yup, grows quickly (although, not that quickly either, compared to $x^n$ or something like that).

                                                                                    • But, danger!, there are also growing numbers of terms in the nominator.


                                                                                    The second line is the important one; this immediately tells me that I cannot trust any further intuition (i.e., random ideas shooting spontaneously through my mind) anymore, and that it is time to find a pen and some paper.



                                                                                    Full disclosure: I am more at home in CS than maths, but the same is true for O notation in CS, it can be quite misleading and unintuitive. Or rather, deceivingly simple and provoking intuitive errors.






                                                                                    share|cite|improve this answer












                                                                                    Seeing such a term, my intuition would be:




                                                                                    • $n²$ in the denominator, yup, grows quickly (although, not that quickly either, compared to $x^n$ or something like that).

                                                                                    • But, danger!, there are also growing numbers of terms in the nominator.


                                                                                    The second line is the important one; this immediately tells me that I cannot trust any further intuition (i.e., random ideas shooting spontaneously through my mind) anymore, and that it is time to find a pen and some paper.



                                                                                    Full disclosure: I am more at home in CS than maths, but the same is true for O notation in CS, it can be quite misleading and unintuitive. Or rather, deceivingly simple and provoking intuitive errors.







                                                                                    share|cite|improve this answer












                                                                                    share|cite|improve this answer



                                                                                    share|cite|improve this answer










                                                                                    answered Jul 28 '17 at 11:20









                                                                                    AnoE

                                                                                    63529




                                                                                    63529






















                                                                                        up vote
                                                                                        3
                                                                                        down vote













                                                                                        "grows much faster than the numerator": you are disregarding the fact that the numerator is actually



                                                                                        $$1+2+3+cdots n, $$ which doesn't grow slower than $n^2$.






                                                                                        share|cite|improve this answer

























                                                                                          up vote
                                                                                          3
                                                                                          down vote













                                                                                          "grows much faster than the numerator": you are disregarding the fact that the numerator is actually



                                                                                          $$1+2+3+cdots n, $$ which doesn't grow slower than $n^2$.






                                                                                          share|cite|improve this answer























                                                                                            up vote
                                                                                            3
                                                                                            down vote










                                                                                            up vote
                                                                                            3
                                                                                            down vote









                                                                                            "grows much faster than the numerator": you are disregarding the fact that the numerator is actually



                                                                                            $$1+2+3+cdots n, $$ which doesn't grow slower than $n^2$.






                                                                                            share|cite|improve this answer












                                                                                            "grows much faster than the numerator": you are disregarding the fact that the numerator is actually



                                                                                            $$1+2+3+cdots n, $$ which doesn't grow slower than $n^2$.







                                                                                            share|cite|improve this answer












                                                                                            share|cite|improve this answer



                                                                                            share|cite|improve this answer










                                                                                            answered Jul 28 '17 at 11:30









                                                                                            Yves Daoust

                                                                                            123k668219




                                                                                            123k668219






















                                                                                                up vote
                                                                                                0
                                                                                                down vote













                                                                                                Very intuitively, there are a lot of positive terms inside the brackets, so they can add up to something non-zero. An extremely loose way to get at what is going on is to over estimate by $n$ times the largest term and under estimate $n/2$ times the middle term, so you have



                                                                                                $$
                                                                                                frac{1}{4} = frac{n}{2}cdot frac{n/2}{n^2} le frac{1}{n^2} + frac{2}{n^2} + cdots + frac{n-1}{n^2} + frac{n}{n^2} le ncdot frac{n}{n^2} = 1
                                                                                                $$



                                                                                                In fact, the answer is just the limit of $n^{-2}binom{n+1}{2} = frac{1}{2}$ as $ntoinfty$, but the above is why you should guess that the answer isn't zero.






                                                                                                share|cite|improve this answer

























                                                                                                  up vote
                                                                                                  0
                                                                                                  down vote













                                                                                                  Very intuitively, there are a lot of positive terms inside the brackets, so they can add up to something non-zero. An extremely loose way to get at what is going on is to over estimate by $n$ times the largest term and under estimate $n/2$ times the middle term, so you have



                                                                                                  $$
                                                                                                  frac{1}{4} = frac{n}{2}cdot frac{n/2}{n^2} le frac{1}{n^2} + frac{2}{n^2} + cdots + frac{n-1}{n^2} + frac{n}{n^2} le ncdot frac{n}{n^2} = 1
                                                                                                  $$



                                                                                                  In fact, the answer is just the limit of $n^{-2}binom{n+1}{2} = frac{1}{2}$ as $ntoinfty$, but the above is why you should guess that the answer isn't zero.






                                                                                                  share|cite|improve this answer























                                                                                                    up vote
                                                                                                    0
                                                                                                    down vote










                                                                                                    up vote
                                                                                                    0
                                                                                                    down vote









                                                                                                    Very intuitively, there are a lot of positive terms inside the brackets, so they can add up to something non-zero. An extremely loose way to get at what is going on is to over estimate by $n$ times the largest term and under estimate $n/2$ times the middle term, so you have



                                                                                                    $$
                                                                                                    frac{1}{4} = frac{n}{2}cdot frac{n/2}{n^2} le frac{1}{n^2} + frac{2}{n^2} + cdots + frac{n-1}{n^2} + frac{n}{n^2} le ncdot frac{n}{n^2} = 1
                                                                                                    $$



                                                                                                    In fact, the answer is just the limit of $n^{-2}binom{n+1}{2} = frac{1}{2}$ as $ntoinfty$, but the above is why you should guess that the answer isn't zero.






                                                                                                    share|cite|improve this answer












                                                                                                    Very intuitively, there are a lot of positive terms inside the brackets, so they can add up to something non-zero. An extremely loose way to get at what is going on is to over estimate by $n$ times the largest term and under estimate $n/2$ times the middle term, so you have



                                                                                                    $$
                                                                                                    frac{1}{4} = frac{n}{2}cdot frac{n/2}{n^2} le frac{1}{n^2} + frac{2}{n^2} + cdots + frac{n-1}{n^2} + frac{n}{n^2} le ncdot frac{n}{n^2} = 1
                                                                                                    $$



                                                                                                    In fact, the answer is just the limit of $n^{-2}binom{n+1}{2} = frac{1}{2}$ as $ntoinfty$, but the above is why you should guess that the answer isn't zero.







                                                                                                    share|cite|improve this answer












                                                                                                    share|cite|improve this answer



                                                                                                    share|cite|improve this answer










                                                                                                    answered Dec 6 '17 at 14:17









                                                                                                    Louis

                                                                                                    859611




                                                                                                    859611






















                                                                                                        up vote
                                                                                                        0
                                                                                                        down vote













                                                                                                        Here's an explanation based on the intuitively obvious fact that a triangle with very little height can nonetheless have substantial area provided it has a very long base.



                                                                                                        Draw the step function $f_n(x)={lceil xrceilover n^2}$ for $0le xle n$, and observe that it lies above the triangle with vertices at $(0,0)$, $(n,0)$ and $(n,{1over n})$ -- i.e., a right triangle with base of length $n$ and height $1over n$, hence area $1over2$. Since the area beneath the step function is $S_n={1over n^2}+{2over n^2}+cdots+{nover n^2}$, this shows that $lim_{ntoinfty}S_n$, if it exists, cannot be less than $1over2$.



                                                                                                        If you like, you can also draw the step function $g_n(x)={lfloor xrfloorover n^2}$ for $0le xle n+1$ which lies below a right triangle with base of length $n+1$ and height $n+1over n^2$ and argue that the area beneath the step function, ${0over n^2}+{1over n^2}+{2over n^2}+cdots+{nover n^2}$, is less than ${1over2}left(n+1over nright)^2$. Combining this with the result of the first paragraph leads to the limit $S_nto{1over2}$.






                                                                                                        share|cite|improve this answer

























                                                                                                          up vote
                                                                                                          0
                                                                                                          down vote













                                                                                                          Here's an explanation based on the intuitively obvious fact that a triangle with very little height can nonetheless have substantial area provided it has a very long base.



                                                                                                          Draw the step function $f_n(x)={lceil xrceilover n^2}$ for $0le xle n$, and observe that it lies above the triangle with vertices at $(0,0)$, $(n,0)$ and $(n,{1over n})$ -- i.e., a right triangle with base of length $n$ and height $1over n$, hence area $1over2$. Since the area beneath the step function is $S_n={1over n^2}+{2over n^2}+cdots+{nover n^2}$, this shows that $lim_{ntoinfty}S_n$, if it exists, cannot be less than $1over2$.



                                                                                                          If you like, you can also draw the step function $g_n(x)={lfloor xrfloorover n^2}$ for $0le xle n+1$ which lies below a right triangle with base of length $n+1$ and height $n+1over n^2$ and argue that the area beneath the step function, ${0over n^2}+{1over n^2}+{2over n^2}+cdots+{nover n^2}$, is less than ${1over2}left(n+1over nright)^2$. Combining this with the result of the first paragraph leads to the limit $S_nto{1over2}$.






                                                                                                          share|cite|improve this answer























                                                                                                            up vote
                                                                                                            0
                                                                                                            down vote










                                                                                                            up vote
                                                                                                            0
                                                                                                            down vote









                                                                                                            Here's an explanation based on the intuitively obvious fact that a triangle with very little height can nonetheless have substantial area provided it has a very long base.



                                                                                                            Draw the step function $f_n(x)={lceil xrceilover n^2}$ for $0le xle n$, and observe that it lies above the triangle with vertices at $(0,0)$, $(n,0)$ and $(n,{1over n})$ -- i.e., a right triangle with base of length $n$ and height $1over n$, hence area $1over2$. Since the area beneath the step function is $S_n={1over n^2}+{2over n^2}+cdots+{nover n^2}$, this shows that $lim_{ntoinfty}S_n$, if it exists, cannot be less than $1over2$.



                                                                                                            If you like, you can also draw the step function $g_n(x)={lfloor xrfloorover n^2}$ for $0le xle n+1$ which lies below a right triangle with base of length $n+1$ and height $n+1over n^2$ and argue that the area beneath the step function, ${0over n^2}+{1over n^2}+{2over n^2}+cdots+{nover n^2}$, is less than ${1over2}left(n+1over nright)^2$. Combining this with the result of the first paragraph leads to the limit $S_nto{1over2}$.






                                                                                                            share|cite|improve this answer












                                                                                                            Here's an explanation based on the intuitively obvious fact that a triangle with very little height can nonetheless have substantial area provided it has a very long base.



                                                                                                            Draw the step function $f_n(x)={lceil xrceilover n^2}$ for $0le xle n$, and observe that it lies above the triangle with vertices at $(0,0)$, $(n,0)$ and $(n,{1over n})$ -- i.e., a right triangle with base of length $n$ and height $1over n$, hence area $1over2$. Since the area beneath the step function is $S_n={1over n^2}+{2over n^2}+cdots+{nover n^2}$, this shows that $lim_{ntoinfty}S_n$, if it exists, cannot be less than $1over2$.



                                                                                                            If you like, you can also draw the step function $g_n(x)={lfloor xrfloorover n^2}$ for $0le xle n+1$ which lies below a right triangle with base of length $n+1$ and height $n+1over n^2$ and argue that the area beneath the step function, ${0over n^2}+{1over n^2}+{2over n^2}+cdots+{nover n^2}$, is less than ${1over2}left(n+1over nright)^2$. Combining this with the result of the first paragraph leads to the limit $S_nto{1over2}$.







                                                                                                            share|cite|improve this answer












                                                                                                            share|cite|improve this answer



                                                                                                            share|cite|improve this answer










                                                                                                            answered Dec 6 '17 at 15:30









                                                                                                            Barry Cipra

                                                                                                            58.7k653122




                                                                                                            58.7k653122






























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