I'm trying to prove that $\sum_{n=-\infty}^\infty\frac{1}{(z-n)^2}$ converges uniformly on any (open) disc that does not contain any integer. Need to prove that $\sum_{n=0}^\infty\frac{1}{(z-n)^2}$ and $\sum_{n-=\infty}^{-1}\frac{1}{(z-n)^2}$ converge in the above sense.
First, let's consider the first series. Let $D(a,r)$ be the open disc centered at $a$ of radius $r$. I guess I need to apply the Weierstrass $M$-test, but I don't know how. I certainly need the triangle inequality: if $z\in D(a,r)$, then $d(z,n)\ge d(a,n)-d(z,a)> d(a,n)-r$ where $n$ is any integer (or in this case any natural number, including zero). Then $\frac{1}{(z-n)^2} <\frac{1}{(d(a,n)-r)^2}$, but this doesn't actually give anything. I guees I need to make the last expression less than $1/n^2$, but how?
For the second sum, I think everything should be analogous, but Why does the Weierstrass $M$-test can be applied to sums from $-\infty$ to $-1$? (And to begin with, what does it mean for such a sum to be convergent? It cannot be written in the usual form $\sum_{n=1}^\infty\frac{1}{(z+n)^2}$, can it?)