Show that $\sum_{r=1}^\infty \frac{1}{(r-z)^2}$ is holomorphic on $\mathbb{C}\backslash\mathbb{N}$

Question

I'm asked to show that $\displaystyle\sum_{r=1}^\infty \dfrac{1}{(r-z)^2}$ is holomorphic on $\mathbb{C}\backslash\mathbb{N}$, and am given the hint that for any $z \in \mathbb{C}\backslash\mathbb{N}$ the series is uniform on some neighbourhood of $z$.

Someone has asked the same question here:

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

but I haven't been shown the result that the answer uses. Does anyone know a better way to do it?

I know I can use Morera's theorem to show that $\displaystyle\int_{\gamma} \sum_{r=1}^\infty \dfrac{1}{(r-z)^2} \ \text{dz} = 0 $ for any closed path $\gamma$, but I'm not sure how to justify exchanging the integral and the sum for any arbitrary path, when the convergence is only uniform on some neighbourhood of each point - not necessarily along the whole arbitrary closed path.

Thanks in advance.

Answer 1

You can use the following simpler fact. If

• $f_n$ are holomorphic in $\Omega$,
• $f_n\to f$ uniformly in $\Omega$,
• $f_n'\to g$ uniformly in $\Omega$,

then $f$ is holomorphic with $f'=g$.

Then check that with $f_n$ equal to the partial sum and $\Omega\Subset\mathbb C\setminus\mathbb N$ the assumptions are verified.