Complex proof (with sum) How can I prove $$\sum_{n=-\infty}^∞\frac{1}{(a+bn)^2}=\frac{π^2}{b^2}  \csc^2 \frac{πa}{b}$$
When $\frac{a}{b}\notin$*Z*?

You may do so with the residue theorem.  Applied to infinite sums, it reads
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \text{Res}_{z=z_k} \pi \cot{\pi z}\, f(z)$$
where $z_k$ are the poles of $f$ in the complex plane outside of the integers.  In your case, there is one pole of $f$, albeit a double pole, at $z=-a/b$.  Then
$$\begin{align}\sum_{n=-\infty}^{\infty} \frac{1}{(a+b n)^2}&= \frac{1}{b^2} \sum_{n=-\infty}^{\infty} \frac{1}{(n + (a/b))^2}\\ &=-\frac{1}{b^2} \text{Res}_{z=-a/b} \frac{\pi \cot{\pi z}}{(z + (a/b))^2}\\ &= -\frac{1}{b^2} \lim_{z \rightarrow -a/b} \frac{d}{dz} \left [(z + (a/b))^2 \frac{\pi \cot{\pi z}}{(z + (a/b))^2}  \right ]  \\ &= -\frac{\pi}{b^2} \left[\frac{d}{dz} \cot{\pi z} \right ]_{z=-a/b}\\ &= \frac{\pi^2}{b^2} \csc^2{\pi \frac{a}{b}} \end{align}$$
as as to be shown.
