A double sum $\sum \limits_{n=1}^{n=\infty}\left(\sum \limits_{k=n}^{k=n^2}\frac{1}{k^2}\right)$

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

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,

$$\sum_{k=1}^\infty\left(\frac1k-\frac1{k^{3/2}}\right)$$

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 $x=1$:

$$\lim_{h\to0}\frac{\zeta(1-h)+\zeta(1+h)}2=\gamma$$

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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