Is this function continuous at all points where the series converges? The function is $$f(x)=\sum_{n=1}^{\infty} \frac{1}{n(1+nx^2)}, x \in \mathbb{R} $$ I want to find out whether this series is continuous at all points where the series converges, any ideas on what theorem/method would provide me with the answer? Thanks :)
 A: We will use a classical $\delta-\epsilon$ proof to show that the series of interest is continuous for $x>0$.  To that end we proceed.
Let $f(x)$ be given by the series
$$f(x)=\sum_{n=1}^\infty\frac{1}{n(1+nx^2)}$$
Restrict $a>0$ and $x$ such that $a/2<x<3a/2$.  Then, we have the estimate 
$$\begin{align}
|f(x)-f(a)|&=\sum_{n=1}^\infty\left|\frac{1}{n(1+nx^2)}-\frac{1}{n(1+na^2)}\right|\\\\
&=|x^2-a^2|\sum_{n=1}^\infty \frac{1}{(1+nx^2)(1+na^2)}\\\\
&\le |x^2-a^2|\frac{4}{a^4}\sum_{n=1}^\infty \frac{1}{n^2}\\\\
&= |x-a|\frac{5 \pi^2}{3a^3}\\\\
&<\epsilon
\end{align}$$
whenever $|x-a|<\delta=\min\left(a/2,\frac{3a^3}{5\pi^2}\,\epsilon\right)$.  And we are done!
A: First, note that the series diverges for $x=0$ since this is just the harmonic series.
Now we look at the interval $ (=\infty,-a)\cup(a,\infty)$ where $a>0$
Notice that if $x\in(-\infty,-a)\cup(a,\infty)$
$\frac{1}{n(1+nx^2)}\leq\frac{1}{n^2x^2}\leq\frac{1}{a^2n^2}$
and $\sum_{n=1}^{\infty}\frac{1}{a^2n^2}$ converges.
Therefore $\sum_{n=1}^{\infty}\frac{1}{n(1+nx^2)}$ converges uniformly on $ (-\infty,-a)\cup(a,\infty)$ 
Since $\frac{1}{n(1+nx^2)}$ is continuous where it's defiend, it follows that $f(x)$ is also continuous on $ (-\infty,-a)\cup(a,\infty)$ 
Since $a>0$ was arbitrary, $f$ is in fact continuous everywhere but at $x=0$ (This especially means it's continuous where it converges)
