Show that $f(z)=\sum_{n=-\infty}^\infty \frac{1}{(n-z)^2}$ converges uniformly on compact sets in $\Bbb C - \Bbb Z$. Show that 

$$f(z)=\sum_{n=-\infty}^\infty \frac{1}{(n-z)^2}$$

converges uniformly on compact sets in $\Bbb C - \Bbb Z$, and is a continuous function there.

Here is the original dutch version:



My first idea is to write this like:

$$1/z^2+\sum_{n=1}^\infty 1/(n-z)^2+1/(n+z)^2$$

And then use the Weisterstraas M test. But this didn't work.
Any other ideas ?
Any compact set $K$ in $\mathbb C\setminus\mathbb Z$ is contained in some disk $B(0,N)$, $N\in\mathbb N$. At the same time $K$ has a positive minimal distance from $\mathbb Z$, $0&lt;\delta=d(K, \mathbb Z)$, since $d(x,\mathbb Z)=\min_{n\in\mathbb Z}|x-n|$ is a continuous function. 

Then for any $z\in K$
\begin{align}
\left|\frac{1}{(z-n)^2}\right|&amp;&lt;\frac{1}{δ^2}&amp;&amp;\text{ for }|n|\le N\\
\left|\frac{1}{(z-n)^2}\right|&amp;&lt;\frac{1}{(|n|-|N|)^2}&amp;&amp;\text{ for }|n|\ge N+1
\end{align}
so that
$$
\sum_{n\in \mathbb Z}\left|\frac{1}{(z-n)^2}\right|\le \frac{2N+1}{δ^2}+2\cdot\frac{\pi^2}6
$$
By the Weierstraß M test convergence is uniform on $K$.



Which means that the idea in the question was right, one just needs to take out a larger middle part.
i think (by taking euler product) that the function is about

$$ \frac{d^{2}}{dx^{2}}log(sin(z)/z) $$

THE infinite Euler product is equal to $ sin(z) $ up to some constant