Prove that $f(x)=\sum^{\infty}_{n=1}\frac{1}{n^2-x^2}$ is continuous at all $x \notin \Bbb N$.
An attempt:
We should consider showing that $\sum^{\infty}_{n=1}\frac{1}{n^2-x^2}$ converges uniformly.
Also,
$$f(x)=\sum^{\infty}_{n=1}\frac{1}{n^2-x^2}=\sum^{\infty}_{n=1}\frac{1}{2n}\left(\frac{1}{n-x}+\frac{1}{n+x}\right)$$
Let $k\in \mathbb Z$ and $a,b\in\mathbb R$ such that $a<b$ and $[a,b]\subset (k,k+1)$ then: $$\frac{1}{|n^2-x^2|}\leq c_n=\max\left(\frac{1}{|n^2-a^2|},\frac{1}{|n^2-b^2|}\right)\quad\forall\ x\in [a,b]$$ and the series $$\sum_n c_n$$ is convergent since $c_n\sim_\infty \frac{1}{n^2}$ so the given series is uniformly convergent in every interval $[a,b]$ and with the continuity of the functions $x\mapsto \frac{1}{n^2-x^2}$ we can conclude.