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 $?
 A: Apart from the first few summands, we have $n^2-y^2>\frac12n^2$ for all $y\approx x$, hence the tail is (uniformly near $x$) bounded by $2\sum_{n>N}\frac1{n^2}$.
A: Let $k\in\mathbb{Z}$ and $a,b\in\mathbb{R}$ such that $k<a<b<k+1$ then we have:
$$\forall x\in[a,b],\quad\frac{1}{|n^2-x^2|}\leq \max\left(\frac{1}{|n^2-a^2|},\frac{1}{|n^2-b^2|}\right)\sim_\infty\frac{1}{n^2}$$
so the series is normal convergent on the interval $[a,b]$ and hence it's uniform convergent on $[a,b]$.
A: 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 $B(x,\delta)$.
A: Another approach to show the series converges absolutely. By the limit comparison test:
$$\forall\,n\in\Bbb N\,,\,n>|x|\;: \lim_{n\to\infty}\frac{\frac{1}{|n^2-x^2|}}{\frac{1}{n^2}}=\lim_{n\to\infty}\frac{n^2}{n^2-x^2}=1$$
Since only a finite number of indexes $\,n\,$ don't fulfil $\,n>|x|\,$ we're done.
