[Math] Showing $\sum_{r=1}^\infty \frac{1}{(r-z)^2}$ is holomorphic on $\mathbb{C}\setminus\mathbb{N}$

Question

Ok, so as per the question title I'm wanting to show that
$\displaystyle\sum_{r=1}^\infty \dfrac{1}{(r-z)^2}$ is holomorphic on $\mathbb{C}\setminus\mathbb{N}$.

I'm thinking that if I show that

$$\int_{\gamma} \sum_{r=1}^\infty \frac{1}{(r-z)^2} \ \text{d}z = 0 $$

Then by Morera's theorem, the sum is holomorphic. The issue is that the series' convergence isn't uniform on all of $\mathbb{C}\setminus\mathbb{N}$. Although the question does hint that for any $z \in \mathbb{C}\setminus\mathbb{N}$ the series is uniform on some neighbourhood of $z$

Any help greatly appreciated!

Answer 1

The holomorphicity of the sum is a direct consequence of Weierstrass' theorem:

Suppose that $f_n(z)$ is holomorphic in the region $\Omega$, and that the sequence $\{f_n(z)\}$ converges to a limit function $f(z)$ in $\Omega$, uniformly on every compact subset of $\Omega$. Then $f(z)$ is holomorphic in $\Omega$. Moreover, $f_n'(z)$ converges uniformly to $f' (z)$ on every compact subset of $\Omega$.