= Solution
Fix any bounded disk avoiding the integers, and choose $R$ with $|z|\leq R$ on it. For sufficiently large $|n|$, say $|n|\geq2R$ and $|n|\geq1$,
$$
|n^2-z^2|\geq n^2-|z|^2\geq\frac34 n^2,
\qquad
\left|\frac1{n^2-z^2}\right|\leq\frac4{3n^2}.
$$
The convergent numerical series $\sum_{n\ne0}n^{-2}$ and the <Weierstrass M-test> give absolute and <uniform convergence> of the tails. The finitely many remaining summands are <holomorphic> away from their integer poles. Thus the partial sums converge locally uniformly on $\mathbb C\setminus\mathbb Z$.
By part (a)(iii), \b[the sum is a <holomorphic function> on $\mathbb C\setminus\mathbb Z$]. The term $n=0$ is $-1/z^2$ and is harmless on this domain; it must not be omitted.
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