In Euler's Introductio in Analysin Infinitorum I found $$S=\sum_{n=0}^\infty \left(\frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}\right)=\frac{4\pi^3}{81\sqrt{3}}.$$ Instead of using Euler's approach, I thought that this sum could be tackled by writing $$S=\sum_{n=-\infty}^\infty \frac{1}{(3n+1)^3}$$ and then using $$\sum_{n=-\infty}^\infty \frac{1}{(z-n)^3}=\pi^3\cot \pi z\operatorname{csc}^2\pi z.$$ However, this poses a problem for me because I actually don't know how to prove that $1/(z-n)^3$ identity. But I do know how to prove analogs of that identity with $1/(z-n)^2$ and $1/(z-n)$.
It led me to the following question: Given that $$\begin{align}\lim_{N\to\infty}\sum_{n=-N}^N\frac{1}{z-n}&=\pi \cot\pi z,\\ \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}&=\pi^2\operatorname{csc}^2\pi z,\\ \sum_{n=-\infty}^\infty \frac{1}{(z-n)^3}&=\pi^3\cot\pi z\operatorname{csc}^2\pi z,\end{align}$$ for all complex $z$ (except for a set of isolated points), does this pattern continue? I.e. can $$\sum_{n=-\infty}^\infty \frac{1}{(z-n)^\alpha}$$ be evaluated as a finite combination of trigonometric functions for every positive integer $\alpha$? If so, how can an arbitrary identity from this family be proved?