Basically, I'm wondering if the following function (defined over all the reals) is continuous:
$f(x)=\sum_{n=1}^{\infty}\frac{1}{n^2+x^2}$
The graph seems to be continuous, but I can't think of any way to prove that the sum itself is continuous.
Edit: Is there a way to do so with the M-test because I haven't covered Hyperbolic Identities
On
Put $$f_N(x)=\sum_{n=1}^{N} \frac{1}{n^2+x^2}$$.
By Weierstrass criterion for uniform convergence we have that $f_N(x)$ converges uniformly on $\mathbb{R}$ because $\frac{1}{n^2+x^2} \le \frac{1}{n^2}$ and $\sum_n \frac{1}{n^2} < \infty$.
We know that $f_N(x)$, $N \ge 1$, are continuous. It follows from the uniform convergence that their limit $\lim_N f_N(x)$ is also continuous, q.e.d.
HINT. Consider the Weierstrass product for the $\sinh$ function: $$ \frac{\sinh z}{z}=\prod_{n=1}^{+\infty}\left(1+\frac{z^2}{\pi^2 n^2}\right) $$ then take logarithmic derivatives of both sides. We get $$ \coth z-\frac{1}{z}=\sum_{n=1}^{+\infty}\frac{2z}{z^2+\pi^2 n^2} $$ By some easy substitutions, we have $$ \sum _{n=0}^{\infty } \frac{1}{n^2+x^2}=\frac{1}{2} \left(\frac{1}{x^2}+\frac{\pi \coth (\pi x)}{x}\right) $$
Can you continue from here?
To use M-test, we have $$ \frac{1}{n^2+x^2} \leq \frac{1}{n^2} $$ Can you continue from here?