= Solution
Let $K\subset\mathbb C\setminus\mathbb Z$ be compact and contained in $|z|\le R$. For $|n|>2R$, $|z-n|\ge|n|/2$, hence $|(z-n)^{-2}|\le4/n^2$ uniformly on $K$. The tail is therefore absolutely and uniformly convergent by the <Weierstrass M-test>. Its finitely many remaining summands are holomorphic on a neighborhood of $K$.
The theorem that a locally uniform limit of holomorphic functions is holomorphic now proves that the sum is \b[analytic on $\mathbb C\setminus\mathbb Z$]. Absolute convergence also permits a shift of the integer index:
$$
f(z+1)=\sum_{n\in\mathbb Z}\frac1{(z+1-n)^2}
=\sum_{m\in\mathbb Z}\frac1{(z-m)^2}=f(z).
$$
Thus the function has period one.
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