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I'm doing a homework problem, and so far I've proved $\sum_{n=-\infty}^\infty \frac{1}{(z+n)^k}=\frac{(-2\pi i)^k}{(k-1)!}\sum_{m=1}^\infty m^{k-1}e^{2\pi imz}$ for $k$ an integer $\geq 2$ and $\text{Im}(z)>0$. The next part of the problem asks me to put $k=2$ and show $\sum_{n=-\infty}^\infty \frac{1}{(z+n)^2}=\frac{\pi^2}{\sin^2(\pi z)}$ for $\text{Im}(z)>0$, but after nearly an hour of bashing I still see it-- I've tried product expansions of sine, using Euler's formula (which equates to showing $\frac{e^{2\pi ir}+e^{-2\pi ir}-2}{16}=-4\pi^2 \sum_{m=1}^\infty me^{2\pi imz}$). Could anybody point out a way to get the above equality from the one I derived? Apologies in advance if I'm just ditzy and it's really obvious.

Also, the next question asks if the above formula is true if $z$ is any complex number that is not an integer, but I'm not really sure I understand what it's asking. Why wouldn't it be true?

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    @user39843 [Here](http://math.stackexchange.com/questions/110494/possibility-to-simplify-sum-limits-k-infty-infty-frac-left)2013-05-25

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Writing $q=e^{2\pi i z}$ we have

$\sum_{m=1}^\infty mq^m=\frac{q}{(1-q)^2}=\frac{1}{(q^{1/2}-q^{-1/2})^2}=\frac{1}{(2i\sin\pi z)^2}.$

Your formula's RHS diverges for $\operatorname{Im}(z)<0$, so one should be suspicious of the corollary formula holding more generally. Hint: consider convergence for real nonintegers and complex conjugation.

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    Thanks! In retrospect it was pretty obvious, as always.2012-03-17