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Compute $\displaystyle\sum _{j=2}^{\infty }\left(\sum _{k=2}^{\infty }\:k^{-j}\right)$

I don't really know where to start. I can see that the inner series is a $p$-series with $p=j$ but where to go after that has me stumped. Any help greatly appreciated.

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    $\begingroup$ Change the order of summation (first on $j$, then on $k$), and rejoice. $\endgroup$
    – Did
    Aug 29, 2017 at 13:03
  • $\begingroup$ If you really want to evaluate the sums in the current order, consider the Maclaurin expansion of the digamma function. $\endgroup$
    – Simply Beautiful Art
    Aug 29, 2017 at 13:07
  • $\begingroup$ Sorry Did still not getting it. Do I calculate the sum of the outer series first? $\endgroup$
    – Eiraus
    Aug 29, 2017 at 13:08
  • $\begingroup$ Sorry Simply Beautiful Art, I have never heard of the digamma function? $\endgroup$
    – Eiraus
    Aug 29, 2017 at 13:09
  • $\begingroup$ Sorry Eiraus, are you asking Beautiful art whether you have never heard of the digamma function! $\endgroup$
    – uniquesolution
    Aug 29, 2017 at 13:11

2 Answers 2

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As Did suggested in the comments, rearrange summation order:

$$\sum\limits_{j=2}^\infty\sum\limits_{k=2}^\infty k^{-j} = \sum\limits_{k=2}^\infty\sum\limits_{j=2}^\infty k^{-j}$$

The inner sum is well-known, since it is a geometric progression with common ratio $\frac{1}{k}$:

$$\sum\limits_{j=2}^\infty k^{-j} = \sum\limits_{j=2}^\infty \frac{1}{k^j} = \frac{1}{k^2}\left(1 + \frac{1}{k} + \frac{1}{k^2} + \frac{1}{k^3} + \dots\right) = \frac{1}{k^2}\frac{1}{1-\frac{1}{k}} = \frac{1}{k^2-k} = \frac{1}{k-1}-\frac{1}{k}$$

Now this is clearly a telescopic series (note how everything except the first summand cancels out):

$$\sum\limits_{k=2}^\infty\left(\frac{1}{k-1}-\frac{1}{k}\right)= \left(\frac{1}{1} - \frac{1}{2}\right)+\left(\frac{1}{2} - \frac{1}{3}\right)+\left(\frac{1}{3} - \frac{1}{4}\right)+\dots = 1 + \left(-\frac{1}{2} + \frac{1}{2}\right) + \left(-\frac{1}{3} + \frac{1}{3}\right) + \dots = 1$$

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    $\begingroup$ You have to justify that you can indeed rearrange the summation order with Fubini's theorem first, I think. $\endgroup$
    – Mariuslp
    Aug 29, 2017 at 13:22
  • $\begingroup$ lisyarus Thank your for the in depth answer and @Mariuslp for the Fubini's theorem tip. I had no idea that changing the order of summation was a 'legal' move. $\endgroup$
    – Eiraus
    Aug 29, 2017 at 13:42
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You can make up the table: $$\begin{array}{c|lccc|r} a_{jk} & \text{} & \text{$k$} & \text{} & \text{} & \mathrm{Total}\\ \hline & \frac{1}{2^2} & \frac{1}{3^2} & \frac{1}{4^2} & \cdots \\ j & \frac{1}{2^3} & \frac{1}{3^3} & \frac{1}{4^3} & \cdots \\ & \frac{1}{2^4} & \frac{1}{3^4} & \frac{1}{4^4} & \cdots \\ & \ \vdots & \vdots & \vdots & \ddots \\ \hline \mathrm{Total} & \frac{1}{2} & \frac{1}{2\cdot 3} & \frac{1}{3\cdot 4} & \cdots & 1 \ \ \ \ \end{array}$$ Note: $$\sum_{m=1}^{\infty} \frac{1}{m(m+1)}=\lim_{n\to\infty} \sum_{m=1}^n \frac{1}{m(m+1)}=\lim_{n\to\infty} \left[\frac12+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}\right]=$$ $$\lim_{n\to\infty} \left[\left(1-\require{cancel}\cancel{\frac12}\right)+\left(\cancel{\frac{1}{2}}-\cancel{\frac13}\right)+\left(\cancel{\frac{1}{3}}-\cancel{\frac14}\right)+\cdots +\left(\cancel{\frac{1}{n}}-\frac{1}{n+1}\right)\right]=$$ $$\lim_{n\to\infty}\left[1-\frac{1}{n+1}\right]=1.$$

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