How to calculate double sum with power?

Question

How to calculate double sum with power like this? $$\sum_{n = 2}^{\infty} \sum_{m = 2}^{\infty} m^{-n}$$ I have tried to search for the series formula equation but could not find one that fits for this. Is it possible to apply the geometric series formula to this one? If yes, how can we do it?

WolframAlpha gives me result $1$, but I don't know how can it be.

sum (sum m^(-n) m=2..infinity) n=2..infinity

Answer 1

Invert the order of the sums, do the geometric sum, partial fractions & telescope \begin{eqnarray*} \sum_{m=2}^{\infty} \sum_{n=2}^{\infty} m^{-n} & =& \sum_{m=2}^{\infty} \frac{1}{m^2\left(1-\frac{1}{m} \right)}\\ & =&\sum_{m=2}^{\infty} \frac{1}{m(m-1)} \\ & =&\sum_{m=2}^{\infty}\left(\frac{1}{(m-1)} -\frac{1}{m}\right) =1. \\ \end{eqnarray*}

Answer 2

By Tonelli, this is $$\sum_{m=2}^\infty \frac{m^{-2}}{1-m^{-1}}=\sum_{m=2}^\infty \frac{1}{m^2-m}=\lim_{t\to\infty}\sum_{k=2}^t \frac1{k^2-k}=\lim_{t\to\infty}\sum_{k=2}^t \frac1{k-1}-\frac1k=\\=\lim_{t\to\infty}1-\frac1t$$