# 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?

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 $$\lim_{x\to\infty}g(x)=0.$$
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.