Why is $\sum_{k=-\infty}^\infty \frac{1}{(z-n)^2}$ uniformly convergent in $|y| \geq 1$ In the Complex Analysis text by Ahlfors', he says that it's easy to see that the series
$$\sum_{k=-\infty}^\infty \frac{1}{(z-k)^2}$$ converges uniformly in the set $\{x+iy:|y| \geq 1\}.$
I can't see why is it the case. I tried using the Weierstrass M-test, but the bounds I got were $\left|\frac{1}{(z-n)^2} \right| \leq 1$ and the series $\sum_{-\infty}^\infty 1$ diverges.
 A: You're right, the series converges only locally uniformly in $\{z : \lvert \Im z\rvert \geqslant 1\}$.
On p. 188, he writes:

It is uniformly convergent on any compact set after omission of the terms which become infinite on the set.

But, in the part in question, Ahlfors doesn't speak about the convergence of the sequence on that set, rather

For $z = x + iy$ we have (Chap. 2, Sec. 3.2, Ex. 4)
  $$\lvert \sin \pi z\rvert^2 = \cosh^2 \pi y - \cos^2 \pi x$$
  and hence $\pi^2/\sin^2\pi z$ tends uniformly to $0$ as $\lvert y\rvert \to \infty$. But it is easy to see that the function $(7)$ has the same property. indeed, the convergence is uniform for $\lvert y\rvert \geqslant 1$, say, and the limit for $\lvert y\rvert \to \infty$ can thus be obtained by taking the limit in each term.

refers to the uniformity - in $y$ - of the convergence for any fixed $x$, or, stronger, for $x$ in any bounded subset of $\mathbb{R}$.
The exposition is indeed not immaculately clear.
A: Lars Ahlfors is right. Let us consider (as usually, $[x]$ denotes the integer part of a real number $x$)$$ \sum_{k=-\infty}^\infty \frac 1 {|z-k|^2 }=\sum_{k=-\infty}^\infty \frac 1 {(x-k)^2+y^2} \le  \sum_{k=-\infty}^\infty \frac 1 {(x-k)^2+1}=$$ $$\sum_{k-[x]=-\infty}^\infty \frac 1 {(x-[x]-k+[x])^2+1}=\sum_{n=-\infty}^\infty \frac 1 {(x-[x]-n)^2+1} \le \sum_{n=1}^\infty \frac 1 {(n-1)^2+1}+ 1+$$  $$\sum_{n=-\infty}^{-1} \frac 1 {(n+1)^2+1} .$$
