Question : Let $f(z)$ be such that along the path $C_N$ (illustrated), $|f(z)| \leq M/{|z|^k}$, where $k > 1$, and $M$ are constants independent of $N$.
Show that $$\sum_{n=-\infty}^{\infty} f(n) = - \big\{ \textrm{sum of the residues of} ~ \pi \cot (\pi z) f(z) ~ \textrm{at the poles of} ~f(z) \big\}.$$
I was able to prove it when $f(z)$ has a finite number of poles, using the Residue theorem.
How can I go by and prove the result when $f(z)$ has an infinite number of poles ? My feeling is this is an application of Mittag-Leffler theorem. But I couldn't really work it out. Any help is much appreciated.
For $f$ meromorphic on the whole complex plane and satisfying your bound then $$0=\lim_{N\to \infty}\frac1{2i\pi}\int_{C_N} f(z)\pi \cot(\pi z)dz=\lim_{N\to \infty} \sum_{a\in Int(C_N)} Res(f(z)\pi \cot(\pi z),a)$$
If $f$ has no poles on $\Bbb{Z}$ and $\sum_n f(n)$ converges then you get your result, where the sum over the poles of $f$ is in the $\lim_{N\to \infty}\sum_{a\in Int(C_N)} $ sense, in general you can't change the order of summation.