For every $n$ in $\mathbb{N}$, let: $$a_{n}=n\sum_{k=n}^{\infty }\frac{1}{k^{2}}$$

Show that the sequence $\left \{ a_{n} \right \}$ is convergent and then calculate its limit.

To prove it is convergent, I was thinking of using theorems like the monotone convergence theorem. Obviously, all the terms $a_{n}$ are positive. So, if I prove that the sequence is decreasing, then by the monotone convergence theorem it follows that the sequence itself is convergent. $a_{n+1}-a_{n}=-\frac{1}{n}+\sum_{k=n+1}^{\infty }\frac{1}{k^{2}}$. But, I can't tell from this that the difference $a_{n+1}-a_{n}$ is negative.

If anybody knows how to solve this problem, please share.

  • $\begingroup$ You could estimate the sum with an integral and then use the Squeeze Theorem. $\endgroup$ Jul 13, 2012 at 14:42
  • $\begingroup$ The limit of the series is computable using Fourier series representation for an appropriate function $f(x)$ and an appropriate substitution for $x$, any of which, unfortunately I can't remember now. $\endgroup$
    – Karthik C
    Jul 13, 2012 at 14:56
  • $\begingroup$ I hesitate to edit the Question in a way that changes your meaning, but shouldn't the next-to-last statement convey that you can't tell "that the difference $a_{n+1} - a_n$ is negative" ? $\endgroup$
    – hardmath
    Jul 13, 2012 at 15:11
  • $\begingroup$ @hardmath: You were absolutely right. I was thinking of negative, but I wrote positive. Typo is fixed $\endgroup$
    – C. Lambda
    Jul 13, 2012 at 15:15
  • $\begingroup$ This limit was also studied in detail at this MSE link. $\endgroup$ Mar 30, 2014 at 20:09

3 Answers 3


[Edit: the first proof or convergence wasn't quite right, so I removed it.]

It is useful to find some estimates first (valid for $n>1$):

$$\sum_{k=n}^{\infty }\frac{1}{k^{2}}<\sum_{k=n}^{\infty }\frac{1}{k(k-1)}=\sum_{k=n}^{\infty }\left(\frac1{k-1}-\frac1k\right)=\frac1{n-1}\\\sum_{k=n}^{\infty }\frac{1}{k^{2}}>\sum_{k=n}^{\infty }\frac{1}{k(k+1)}=\sum_{k=n}^{\infty }\left(\frac1{k}-\frac1{k+1}\right)=\frac1{n}$$

The last equality in each of these lines holds because those are telescoping series.

This gives us the estimate: $1<a_n<\frac{n}{n-1}$. By the squeeze theorem, we can conclude that our sequence converges and $\lim_{n\to\infty}a_n=1$.

  • $\begingroup$ Why is: $m>n$, we have $\frac{m}{(l+m)^2}<\frac{n}{(l+n)^2}$? I can't see it. I had $\frac{m}{(l+m)^2}-\frac{n}{(l+n)^2}=\frac{(m-n)(l^{2}-mn)}{(l+m)^{2}(l+n)^{2}}$ which depends on the sign of $(l^{2}-mn)$ $\endgroup$
    – C. Lambda
    Jul 13, 2012 at 15:25
  • $\begingroup$ @C.Lambda: I made a mistake, sorry. Luckily, we don't that calculation to solve the problem. $\endgroup$
    – Dejan Govc
    Jul 13, 2012 at 15:27
  • $\begingroup$ Oops, that's a typo; "don't that" should be "don't need that". $\endgroup$
    – Dejan Govc
    Jul 13, 2012 at 15:42
  • $\begingroup$ I like the trick you used to prove the convergence. Thanks a lot for sharing... $\endgroup$
    – C. Lambda
    Jul 13, 2012 at 15:45
  • $\begingroup$ +1: Nice use of telescoping and Squeeze Theorem! Far simpler than the approach I had been considering.... $\endgroup$ Jul 13, 2012 at 16:36

$$ n\sum_{k=n}^\infty\frac1{k^2}=\sum_{k=n}^\infty\frac{n^2}{k^2}\frac1n\tag{1} $$ is a Riemann sum ($x=\frac{k}{n}$ and $\mathrm{d}x=\frac1n$) for $$ \int_1^\infty\frac1{x^2}\mathrm{d}x=1\tag{2} $$ Although $(2)$ is an improper Riemann integral, because $\frac1{x^2}$ is decreasing, we can still use the rectangular upper and lower approximations to $(2)$: $$ \sum_{k=n+1}^\infty\frac{n^2}{k^2}\frac1n<\int_1^\infty\frac1{x^2}\mathrm{d}x< \sum_{k=n}^\infty\frac{n^2}{k^2}\frac1n\tag{3} $$ where the sums in $(3)$ differ by $\frac1n$. Therefore, combining $(1)$-$(3)$ yields $$ 1<n\sum_{k=n}^\infty\frac{1}{k^2}<1+\frac1n\tag{4} $$ which by the sandwich theorem gives $$ \lim_{n\to\infty}n\sum_{k=n}^\infty\frac{1}{k^2}=1\tag{5} $$

  • $\begingroup$ This is really a very nice proof! I just want to make sure of something: If I am not mistaken, your proof takes care of both parts at the same time (i.e, part1: prove it is convergent, part2: find its limit). Is that right? $\endgroup$
    – C. Lambda
    Jul 13, 2012 at 15:43
  • 1
    $\begingroup$ This approach can be difficult to justify, as the usual "the Riemann sum will tend to the integral" is only certain for proper Riemann integrals, while this one is over an infinite range. In practice if I want to find a value this is what I do as well, but when I have to rigorously show the result then I don't know how to fix this. $\endgroup$ Jul 13, 2012 at 15:47
  • $\begingroup$ +1 Simple and beautiful...damn, two out of the first two answers in this thread! $\endgroup$
    – DonAntonio
    Jul 13, 2012 at 16:08
  • $\begingroup$ @RagibZaman: Because $\dfrac1{x^2}$ is monotonically decreasing, we can use rectangular approximations to get $$ \sum_{k=n+1}^\infty\frac{n^2}{k^2}\frac1n<\int_1^\infty\frac1{x^2}\mathrm{d}x< \sum_{k=n}^\infty\frac{n^2}{k^2}\frac1n $$ The difference between the left and right sums is $\dfrac1n$, so we get $$ 1<\sum_{k=n}^\infty\frac{n^2}{k^2}\frac1n<1+\frac1n $$ and we have a sandwich limit. $\endgroup$
    – robjohn
    Jul 13, 2012 at 20:03
  • $\begingroup$ @C.Lambda: Ragib Zaman raised a point about the improper Riemann integral that is easily handled in my previous comment. $\endgroup$
    – robjohn
    Jul 13, 2012 at 20:08

This series, (set the first term aside for now) can be pictured as rectangles of width $1$ and heights $\frac1{n^2}$ placed side to side.

The sum of the areas of the rectangle is the integral of the function $\displaystyle\frac1{\lceil x\rceil^2}$ over $(1,\infty)$. Which is clearly bounded by the integral of $\displaystyle\frac1{x^2}$ over $(1,\infty)$ which in turn is bounded. Now add back the first term. Boundedness remains.

(Similarly you can find an integral which binds it below.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .