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How to evaluate $\displaystyle\sum_{n=1}^{n=\infty}\left(\sum_{k=n}^{k=n^2}\frac{1}{k^2}\right)$?

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This question has generated three good and fairly unique answers, so far (+1). –  robjohn Nov 4 '11 at 20:37

4 Answers 4

up vote 10 down vote accepted

The sum diverges. To see this, lower bound the inner summation by a telescoping sum by writing $\frac{1}{k^2} > \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$. Now use the fact that the harmonic series diverges.

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$$\begin{align*} \sum_{n=1}^{n=\infty}\left(\sum_{k=n}^{k=n^2}\frac{1}{k^2}\right) &= \sum_{k=1}^\infty\;\sum_{n=\lceil\sqrt{k}\rceil}^k\frac1{k^2}\\ &=\sum_{k=1}^\infty\frac{k-\lceil\sqrt{k}\rceil+1}{k^2}\\ &\ge \sum_{k=1}^\infty\frac{k-\sqrt{k}}{k^2}\\ &=\sum_{k=1}^\infty\left(\frac1k-\frac1{k^{3/2}}\right), \end{align*}$$

which clearly diverges.

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By counting how many times a particular $k$ appears, we get $$ \sum_{n=1}^{n=\infty}\left(\sum_{k=n}^{k=n^2}\frac{1}{k^2}\right)=\sum_{k=1}^\infty\frac{k-\left\lceil\sqrt{k}\;\right\rceil+1}{k^2}\ge\sum_{k=1}^\infty\frac{1}{2k} $$ which diverges since the harmonic series diverges and $\left\lceil\sqrt{k}\;\right\rceil-1\le k/2$.

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It doesn’t really affect the argument, but you want the ceiling: you need $n^2\ge k$, $n\ge \sqrt{k}$. –  Brian M. Scott Nov 4 '11 at 20:34
    
@Brian: oops, you're right. Thanks, I will fix it. –  robjohn Nov 4 '11 at 20:43

It diverges.

By definition of the polygamma function the inner sum is $\sum_{k=n}^{n^2} \frac{1}{k^2} = \psi^{(1)}(n) - \psi^{(1)}(n^2+1)$.

For large $n$, its asymptotic expansion is: $$ \psi^{(1)}(n) - \psi^{(1)}(n^2+1) \sim \frac{1}{n} - \frac{1}{2 n^2} + o\left( n^{-2} \right) $$ Thus, $\sum_{n=1}^m \left( \psi^{(1)}(n) - \psi^{(1)}(n^2+1) \right) \sim \ln(m) + O(1)$ for large $m$.

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