Convergence of the series $\sum_{n=-\infty}^{\infty} \frac{1}{(z-n)^2}$ Edit: 
On what compact subset of $\mathbb{C}$ does the series $$\sum_{n=-\infty}^{\infty} \frac{1}{(z-n)^2}$$ convergence absolutely(uniformly)? Or we can discuss the convergence of the given series using a suitable approach?
 A: This series converges in many compact sets.
For instance on $\overline{B(0,\frac{1}{2}})$
In general for a given compact set $K$ this series converges uniformly on the set $K \setminus \Bbb{N} $  and it converges also uniformly on every compact set that does not contain a natural number.
A: If you want convergence on some compact subset of $\mathbb C$ take the compact set as $\{0\}$. Convergence on all compact subsets is not meaningful since the series is not even defined on the compact set $\{1\}$
A: FYI, 
$\sum\limits_{n=-\infty}^{\infty} \dfrac{1}{(z-n)^2}=\sum\limits_{n=1}^{\infty} \dfrac{1}{(z-n)^2}+\sum\limits_{n=1}^{\infty} \dfrac{1}{(z+n)^2}+\dfrac{1}{z^2}\tag1$
As $\frac{1}{(z-n)^2}=\frac{d}{dz}(\frac{1}{n-z})\tag2$ 
and $\frac{1}{(z+n)^2}=-\frac{d}{dz}(\frac{1}{n+z})\tag3$ 
Swapping the order of the sums and derivatives we get: 
$\dfrac{d}{dz}\Big[\sum\limits_{n=1}^{\infty} \dfrac{1}{(n-z)}-\sum\limits_{n=1}^{\infty} \dfrac{1}{(z+n)}\Big]+\dfrac{1}{z^2}\tag4$
Known that $\psi(z+1)=-\gamma+\sum\limits_{n=1}^\infty\frac{1}{n}-\sum\limits_{n=1}^\infty\frac{1}{n+z}$ where $\psi$ is the digamma function.
And applying that $\psi(z+1)=\psi(z) +\frac{1}{z}$
we have the followings: 
$\dfrac{d}{dz}\Big[-\psi(1-z)+\psi(z)+\frac{1}{z}\Big]\tag5$
Introducing the reflection formula of digamma function: $\psi(1-z)-\psi(z)=\pi \cot{\pi z}$, performing the derivation  in (5) and put it back into the expression (4)we get: 
$\sum\limits_{n=-\infty}^{\infty} \dfrac{1}{(z-n)^2}=\dfrac{\pi^2}{\sin^2(\pi z)}\tag6$
