For function defined as $$ f(x)=\sum_{n=1}^{\infty}\frac{1}{x^2+n^2} $$ check if $f$ is continuous and differentiable function.
My approach:
I would like to use the connection between this sum and some integral so tried to write that $$ f(x)=\sum_{n=1}^{\infty}\frac{1}{x^2+n^2}=\frac{1}{n}\sum_{n=1}^{\infty}\frac{1/n}{\left(\frac{x}{n}\right)^2+1} $$ but don't know what to do now because of $1/n$ in numerator.
By the Weirstrass-M test, and the bound $$ \frac{1}{x^2+n^2}\leq\frac{1}{n^2} $$ $f(x)$ is continuous on $\mathbb{R}$ as the uniform limit of continuous functions.
For differentiability, we really do the same thing, using the fact that if a sequence of functions $f_N$ is differentiable for all $N$, and $f_N$ and $f_N'$ converge uniformly, then the uniform limit is differentiable.
In our setting, this requires investigating the convergence of $$ f_N'(x)=\frac{d}{dx}\sum_{n=1}^N\frac{1}{x^2+n^2}= \sum_{n=1}^N\frac{-2x}{(x^2+n^2)^2} $$ Which again follows from the Weirstrass M test. Indeed, use calculus to establish $$ \frac{2|x|}{(x^2+n^2)^2}\leq 2\frac{n/\sqrt{3}}{(n^2/3+n^2)^2}=O(1/n^3) $$ And conclude.
edit: as noted by @zhw, since differentiability is local, we may just show uniform convergence on compacts, establishing the bound on the differentiated partial sums for $|x|\leq a$, $$ \left|\frac{-2x}{(x^2+n^2)^2}\right|\leq\frac{2a}{n^4} $$