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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
up vote 11 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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You should note that the series can be summed up using Ramanujan's summation or Cauchy principal value of the Zeta function.

Given the result obtained by Brian M. Scott,


we can see that the second part is $-\zeta(3/2)$.

The first part is the harmonic series.

Harrmonic series has Ramanujan's sum equal to Euler-Mascheroni constant $\gamma$, which is also the Cauchy principal value of $\zeta(x)$ in $x=1$:


As such we can say the generalized sum of this divergent series is $\gamma-\zeta(3/2)=-2.03516...$

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Since $$\sum_{k = n}^{n^2} \frac{1}{k^2} \ge \frac{1}{n^2} + \int_n^{n^2} \frac{dx}{x^2} = \frac{1}{n}$$ and $\sum_{n = 1}^\infty \frac{1}{n}$ diverges, by the comparison test, the series $\sum_{n = 1}^\infty \sum_{k = n}^{n^2} \frac{1}{k}$ diverges.

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