7
$\begingroup$

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.

  • 0
    You could estimate the sum with an integral and then use the Squeeze Theorem.2012-07-13
  • 0
    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.2012-07-13
  • 0
    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*" ?2012-07-13
  • 0
    @hardmath: You were absolutely right. I was thinking of negative, but I wrote positive. Typo is fixed2012-07-13
  • 0
    This limit was also studied in detail at this [MSE link](http://math.stackexchange.com/questions/685435/trying-to-get-a-bound-on-the-tail-of-the-series-for-zeta2).2014-03-30

3 Answers 3

9

[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. By the squeeze theorem, we can conclude that our sequence converges and $\lim_{n\to\infty}a_n=1$.

  • 0
    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)$2012-07-13
  • 0
    @C.Lambda: I made a mistake, sorry. Luckily, we don't that calculation to solve the problem.2012-07-13
  • 0
    Oops, that's a typo; "don't that" should be "don't need that".2012-07-13
  • 0
    I like the trick you used to prove the convergence. Thanks a lot for sharing...2012-07-13
  • 0
    +1: Nice use of telescoping and Squeeze Theorem! Far simpler than the approach I had been considering....2012-07-13
5

$$ n\sum_{k=n}^\infty\frac1{k^2}=\sum_{k=n}^\infty\frac{n^2}{k^2}\frac1n\tag{1} $$ 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

  • 0
    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?2012-07-13
  • 1
    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.2012-07-13
  • 0
    +1 Simple and beautiful...damn, two out of the first two answers in this thread!2012-07-13
  • 0
    @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.2012-07-13
  • 0
    @C.Lambda: Ragib Zaman raised a point about the improper Riemann integral that is easily handled in [my previous comment](http://math.stackexchange.com/questions/170356/proving-convergence-of-a-sequence-and-then-finding-its-limit/170376#comment391674_170376).2012-07-13
0

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.)