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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.

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    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

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[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 $\lim_{n\to\infty}a_n=1$.

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    +1: Nice use of telescoping and Squeeze Theorem! Far simpler than the approach I had been considering....2012-07-13
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$ 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 which by the sandwich theorem gives $ \lim_{n\to\infty}n\sum_{k=n}^\infty\frac{1}{k^2}=1\tag{5} $

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    @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
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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.)