Differentiability of $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{1+xn^{2}}\right)$ on $(0,+\infty)$ Let $$f(x) = \sum_{n=1}^{\infty}\left(\frac{1}{1+xn^2}\right)$$ Show that $f(x)$ is differentiable on $(0, +\infty)$.
 A: First of all, the series converges for any positive $x$, since $1/(1+xn^2)\le 1/xn^2$ and the sum of the inverses of the squares converges. 
The derivative of the $n$-th term is $-(1+xn^2)^{-2}n^2$, so for any closed interval $[a,b]\subset(0,\infty)$ we have that the series of termwise derivatives is bounded as follows: 
 $$ \left|\sum_n-\frac{n^2}{(1+xn^2)^2}\right|\le\sum\frac{n^2}{x^2n^4}=\frac1{x^2}\sum\frac1{n^2}, $$
so the series converges uniformly, as $1/x^2$ is bounded on $[a,b]$. 
It follows that on $[a,b]$, $f$ is the uniform limit of its partial sums, and that the derivative of $f$ exists, and is the series of termwise derivatives. 
Since this holds in any closed subinterval of $(0,\infty)$, the result follows. Note that we are not claiming that any of the convergences is uniform on the whole positive axis. 
The result we are using is

If $(f_n)_n$ is a sequence of differentiable functions on an interval $I$, $a\in I$, $(f_n(a))_n$ converges, and $(f_n')_n$ converges uniformly, then $(f_n)$ converges uniformly to a differentiable function $f$ such that $f_n'\to f'$.

This is a standard result and most analysis books should contain a proof, see for instance Theorem 6.3.1 of Abbott's Understanding analysis.
