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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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This is interesting. $$\eqalign{ & \sum\limits_{n = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{z + n}}} = \frac{\pi }{{\sin \pi z}} \cr & {\left( {\sum\limits_{n = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{z + n}}} } \right)^2} = \sum\limits_{n = - \infty }^\infty {\frac{1}{{{{\left( {z + n} \right)}^2}}}} = \frac{{{\pi ^2}}}{{{{\sin }^2}\pi z}} \cr} $$ Is the formula second formula true? –  Pedro Tamaroff Apr 26 '12 at 23:47
    
nice solution, but how can you show that $$ \sum_{n = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{z + n}}} = \frac{\pi }{{\sin \pi z}}?$$ –  user39843 Mar 18 '13 at 12:39
    
@user39843 Here –  Pedro Tamaroff May 25 '13 at 1:14
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1 Answer 1

up vote 5 down vote accepted

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. –  Jay C Mar 17 '12 at 1:14
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