Uniform convergence of series $\sum_{n=1}^{\infty} \frac{1}{(nx)^2+1}$ Does the following series
$$
g(x) = \sum_{n=1}^{\infty} \frac{1}{(nx)^2+1}
$$
converge uniformly


*

*on $x \in (0,1)$

*on $x\in (1, \infty)$

*Calculate limit $\lim_{x \rightarrow \infty} g(x)$.
My attempts:
when $x \in (1, \infty)$ we have 
$$
\sum_{n=1}^{\infty}  \frac{1}{(nx)^2+1} \leq 
\sum_{n=1}^{\infty} \frac{1}{n^2x^2} \leq
\sum_{n=1}^{\infty} \frac{1}{n^2}
$$
so  series converge uniformly based on Weierstrass Theorem.
when $x \in (0, 1)$ we have
$$
\lim_{n \rightarrow \infty} \frac{1}{(nx)^2+1} = 0 = f(x) \\ 
f_n \rightarrow f \\
\text{We want to ensure that for every }\epsilon\sup_{x \in (0,1)} \left|\frac{1}{(nx)^2+1} - 0 \right | < \epsilon 
$$
Lets find its extreme points by differentiating:
$$
f_n'(x) = \frac{-2n^2x}{((nx)^2 +1)^2}
$$
so the function is strictly decreasing and has may have an extremum when x = 0. But since $x = 0 \notin (0,1)$ we use limit to find values near zero:
$$
\lim_{\epsilon \rightarrow 0^{+}} \frac{1}{(n\epsilon)^2+1} = 1 \gt \epsilon
$$
So series converge uniformly only when $x > 1$. 
Is that a correct reasoning? How can I find its limit? Does it equal the function $f$ it converges to so it's just 0?
 A: I would do it as follows: if your series was uniformly convergent on $(0,1)$, then the sequence $\bigl(f_n(x)\bigr)_{n\in\mathbb N}$, where $f_n(x)=\frac1{(nx)^2+1}$, would converge uniformly to the null function. But it doesn't, since$$(\forall n\in\mathbb N):f_n\left(\frac1n\right)=\frac12.$$
On the other hand, if $N\in\mathbb N$, then$$g(N)=\sum_{n=1}^\infty\frac1{(Nn)^2+1}<\sum_{n=1}^\infty\frac1{(Nn)^2}<\sum_{n=N}^\infty\frac1{n^2}\to_{N\to\infty}0.$$This, together with the fact that $g$ is decreasing on $(1,\infty)$ (since its the sum of decreasing functions), shows that $\lim_{x\to\infty}g(x)=0.$
A: If you like an overkill, you may notice that $\frac{1}{n^2+a^2}=\int_{0}^{+\infty}\frac{\sin(nt)}{n}e^{-at}\,dt$ for any $a,n\in\mathbb{N}^+$.
It follows that
$$ \sum_{n\geq 1}\frac{1}{n^2 x^2+1} = \frac{1}{x^2}\sum_{n\geq 1}\int_{0}^{+\infty}\frac{\sin(nt)}{t} e^{-t/x}\,dt =\frac{1}{x^2}\int_{0}^{+\infty}W(t)e^{-t/x}\,dt$$
where $W(t)$ is the $2\pi$-periodic extension of the function (sawtooth wave) which equals $\frac{\pi-t}{2}$ on $(0,2\pi)$. 
By explicit integration, it follows that
$$ \sum_{n\geq 1}\frac{1}{n^2 x^2+1} = \frac{\pi\coth\frac{\pi}{x}-x}{2x} $$
so the answer to the third point is simply zero. The RHS is unbounded in a right neighbourhood of the origin, hence we cannot have uniform convergence on $(0,1)$. If $x\in(0,1)$ the RHS is actually extremely close to $\frac{\pi-x}{2x}$. On the other hand the uniform convergence over $[1,+\infty)$ is almost trivial since the LHS is a positive, continuous and decreasing function of the $x$ variable.
