Properties of the function defined by $g(x) = \sum\limits_{n=0}^{\infty} \frac{1}{1+n^2x^2}$ I am looking at the function $g:\mathbb{R} \rightarrow \mathbb{R}$ defined as $$g(x) = \sum\limits_{n=0}^{\infty} \frac{1}{1+n^2x^2}$$
I would like to know if this function is convergent, continuous and differentiable.
For convergence I have $g(0)$ is divergent because we get a sum of 1, but for $x \not = 0$ we get $$0< \sum\limits_{n=0}^{\infty} \frac{1}{1+n^2x^2} < \sum\limits_{n=0}^{\infty} \frac{1}{n^2x^2} = \frac{1}{x^2} \sum\limits_{n=0}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6x^2}$$ and since $g$ is monotonic and bounded it converges (right?)
Since the pointwise limit doesn't exist for $x \not = 0$ then $g$ doesn't converge uniformly making proving continuity more difficult.
For differentiability I know that if $g_m(x)$ (partial sum of g) is point-wise convergent and $g_m'(x)$ is uniformly convergent then $g'(x) = lim_{m \rightarrow \infty } g'_m(x)$ but again I can't use this due to pointwise limit not existing. 
For any of these properties could I look at $g(x)|_{(0,\infty)}$ and $g(x)|_{(-\infty,0)}$ and make conclusions from this?
Any guidance or solutions would be helpful,
Thanks.
 A: First notice that $g$ is even so it suffices to studied it on the interval $[0,\infty)$. Clearly $g$ isn't defined on $0$ and for $x>0$ we have
$$g_n(x)=\frac1{1+n^2x^2}\sim_\infty\frac{1}{x^2}\frac1{n^2}$$
and since the series 
$$\sum_{n\ge1}\frac1{n^2}$$
is a convergent Riemann series then $g$ is defined on $(0,\infty)$.
Now for the uniform convergence, let $a>0$ then
$$\frac1{1+n^2x^2}\le\frac1{1+n^2a^2},\quad\forall x\ge a$$
then we have the uniform convergence on every interval $[a,+\infty)$ and so the function $g$ is continuous on $(0,+\infty)$. There's not a uniform convergence on $(0,+\infty)$ since
$$\lim_{x\to0}g(x)$$
doesn't exist. We can prove that $g$ is $\mathscr C^\infty$ on $(0,+\infty)$ since we can prove that the series
$$\sum_{n\ge0}g_n^{(k)}(x),\quad \forall k\ge1$$
is uniformly convergent on every compact in $(0,\infty)$.
A: If $|x|\gt r$, we have that
$$
\begin{align}
\left|\,\sum_{n=N}^\infty\frac1{1+n^2x^2}\,\right|
&\le\frac1{r^2}\sum_{n=N}^\infty\frac1{n^2}\\
&\le\frac1{r^2}\sum_{n=N}^\infty\frac1{n(n-1)}\\
&=\frac1{r^2(N-1)}
\end{align}
$$
so the sum converges uniformly outside of any neighborhood of $0$.
Furthermore, using this answer, we get
$$
\begin{align}
\sum_{n=0}^\infty\frac1{1+n^2x^2}
&=\frac1{x^2}\sum_{n=0}^\infty\frac1{\frac1{x^2}+n^2}\\
&=\frac1{2ix}\sum_{n=0}^\infty\left(\frac1{n-\frac ix}-\frac1{n+\frac ix}\right)\\
&=\frac12-\frac1{2ix}\sum_{n\in\mathbb{Z}}\frac1{n+\frac ix}\\
&=\frac12-\frac1{2ix}\pi\cot\left(\pi\frac ix\right)\\
&=\frac12+\frac\pi{2x}\coth\left(\frac\pi{x}\right)
\end{align}
$$
