Convergence of the series $\sum_{n=-\infty}^{\infty} \frac{1}{(z-n)^2}$

Question

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?

Answer 1

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 $\{1\}$

Answer 2

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.

Answer 3

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:

$\sum\limits_{n=-\infty}^{\infty} \dfrac{1}{(z-n)^2}=\dfrac{\pi^2}{\sin^2(\pi z)}\tag6$