Convergence of the infinite series $ \sum_{n = 1}^\infty \frac{1} {n^2 - x^2}$ How can I prove that for every $ x \notin \mathbb Z$ the series
$$ \sum_{n = 1}^\infty \frac{1} {n^2 - x^2}$$
converges uniformly in a neighborhood of $ x $?

If $x \notin \mathbb{Z}$, then there exists some $\delta>0$ such that $(x-\delta,x+\delta) \cap \mathbb{Z} = \emptyset$. Then $s(y) = \sum_{n = 1}^\infty \frac{1} {n^2 - y^2}$ converges uniformly for $y \in B(x,\delta)$.
To see this, let $\epsilon>0$ and choose $N$ such that $N^2 > 2 (|x|+\delta)^2$. Then if $n \ge N$ and $|x-y|< \delta$, we have $n^2-y^2 \ge \frac{1}{2} n^2$. Now choose $N' \ge N$, such that $\sum_{k=N'}^\infty \frac{1}{k^2} < \frac{1}{2} \epsilon$.
Then if $n \ge N'$, we have $0 \le \sum_{k = n}^\infty \frac{1} {k^2 - y^2} \le \sum_{k = N'}^\infty \frac{1} {k^2 - y^2} \le 2\sum_{k = N'}^\infty \frac{1} {k^2} < \epsilon$. Hence the convergence is uniform on $B(x,\delta)$.
