 Solution

For each triangle compactly contained in $U$, part (i) permits the interchange of the series and its boundary integral. Cauchy's integral theorem makes every term zero. The uniform limit is continuous, so Morera's theorem yields that $f$ is holomorphic on $U$.

For the final application, each function $(z-k{)}^{-2}$ is holomorphic away from its pole. Fix a positive integer $n$. For $z\in U_n$ and $k>2n$,The numerical series ${\sum }_{k>2n}4/k^2$ converges, so the Weierstrass M-test gives uniform convergence of the tail on $U_n$. Adding the finitely many initial terms preserves uniform convergence. Those initial terms need not be bounded near their excluded poles; only the tail needs the uniform bound. By the first conclusion, the full sum is holomorphic on each $U_n$. Their union is $\mathbb{C}\setminus \mathbb{N}$, hence