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For an integer $q \geq 2$ and $1 \leq a \leq q/2$, I am asked to simplifty $$ \sum_{m = -\infty}^{\infty} \frac{1 }{ (qm + a)^2} $$ I know that the function involves $1 / \sin$, but I am unable to find the sum.

I am not looking for a direct answer, just a general method to compute series such as these. We know that when $q = 1, a = 0, m = 1, 2, 3, \ldots$ , the sum equals $\pi^2 / 6$, which can be proven using Taylor series expansions, but I was unable to do something similar here.

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  • $\begingroup$ $\displaystyle\frac{\pi^2}{\sin^2(\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ (the LHS minus the RHS is a bounded entire function which vanishes at $\infty$) $\endgroup$ – reuns Nov 19 '17 at 1:52
  • $\begingroup$ Why is it bounded? $\endgroup$ – limitIntegral314 Nov 19 '17 at 2:02
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    $\begingroup$ $\frac{\pi^2}{\sin^2(\pi z)} - \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ is $1$-periodic and analytic and bounded on $\Re(z) \in [0,2]$ $\endgroup$ – reuns Nov 19 '17 at 2:05
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Let us start with the Weierstrass product for the sine function: $$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) $$ and apply $\frac{d^2}{dz^2}\log(\cdot)$ to both sides: $$\frac{\pi^2}{\sin^2(\pi z)}-\frac{1}{z^2} = \sum_{n\geq 1}\left[\frac{1}{(z-n)^2}+\frac{1}{(z+n)^2}\right] $$ then rearrange and multiply both sides by $\frac{1}{q^2}$: $$\sum_{m\in\mathbb{Z}}\frac{1}{(mq-qz)^2} = \frac{\pi^2}{q^2\sin^2(\pi z)}. $$ If we set $z=-\frac{a}{q}$, we get: $$\boxed{\sum_{m\in\mathbb{Z}}\frac{1}{(mq+a)^2} = \color{red}{\frac{\pi^2}{q^2\sin^2\left(\frac{\pi a}{q}\right)}}.}$$

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