# Uniform convergence of $\sum_{n=-\infty}^{\infty}\frac{1}{n^2 - z^2}$ on any disc contained in $\mathbb{C}\setminus\mathbb{Z}$

I'm currently revising some complex analysis, and need to show that the series $$\sum_{n=-\infty}^{\infty}\frac{1}{n^2 - z^2}$$ defines a holomorphic function on $\mathbb{C}\setminus\mathbb{Z}$. The hint that the question gives me is to consider any disc contained in $\mathbb{C}\setminus\mathbb{Z}$ and show that the series converges uniformly there.

I can see how to complete the question once I've done this, but it's late, I'm tired, and I can't figure out how to show the series converges uniformly. I've tried the Weierstrass M-Test, but this failed for me on discs where $z$ was close to $n$, for obvious reasons.

Any hints would be greatly appreciated.

• Something involving Laurent Series/Cauchy Integral formula? Not sure otherwise – Chill2Macht May 21 '16 at 22:51

Let $K \subset (B(0,R) \cap \mathbb{C} \setminus \mathbb{Z})$ be a compact set and set $d(K,\mathbb{Z}) = \delta > 0$. If $|n| > R$, then for all $x \in K$ we have

$$|n^2 - x^2| \geq ||n|^2 - |x|^2| = |n|^2 - |x|^2 \geq |n|^2 - R^2.$$

If $|n| \leq R$ then for all $x \in K$ we have

$$|n^2 - x^2| = |x - n||x - (-n)| \geq \delta^2.$$

Hence,

$$\sup_{x \in K} \left| \frac{1}{n^2 - x^2} \right| \leq \begin{cases} \frac{1}{\delta^2} & |n| \leq R, \\ \frac{1}{|n|^2 - R^2} & |n| > R \end{cases}$$

and so the series converges uniformly on $K$ by the Weierstrass $M$-test. Finish by taking $K = \overline{B(x_0,r)}$ for all $x_0 \in \mathbb{C} \setminus \mathbb{Z}$ and small enough $r$.