Proving $\sum^\infty_{n=1} \frac{1}{n^3} \sin(nx)$ is uniformly continuous

Question

$\sum^\infty_{n=1} \frac{1}{n^3} \sin(nx)$

to prove this, I'm trying to prove $\frac{1}{n^3} \sin(nx)$ is uniformly continuous.

So I let $y = x + \delta$ and did

$\frac{1}{n^3} \sin(nx + n\delta) - \frac{1}{n^3} \sin(nx) < \epsilon$

However I have no idea how to proceed the proof from here.

$\sum^\infty_{n=1} \frac{1}{n^3} \sin(nx)$ is continuous and periodic. — Anne Bauval, Jun 06 '23 at 06:04
@zkutch $\epsilon$ is given. Your upper bound may be larger hence useless. — Anne Bauval, Jun 06 '23 at 06:08
If I show $\frac{1}{n^3} \sin (nx)$ is lipschitz, will it be enough? Or should I show it uniformly converges? If so, how can I show it uniformly converges? — , Jun 06 '23 at 06:08
Duplicate: continuous and Periodic function is Uniformly continuous — Anne Bauval, Jun 06 '23 at 06:13
$\left|\frac{1}{n^3} \sin(nx + n\delta) - \frac{1}{n^3} \sin(nx)\right| =2\frac{1}{n^3}\left|\sin \frac{n\delta}{2}\cos\frac{2nx + n\delta}{2} \right|\leqslant \frac{n\delta}{n^3}$ — zkutch, Jun 06 '23 at 06:22
Can't I use dirichlet test? Isn't $\sum^\infty_{n=1}\sin(nx)$ bounded? — , Jun 06 '23 at 06:22

score 0 · Accepted Answer · answered Jun 06 '23 at 07:09

First, since $|\frac{1}{n^3}sin(nx)|\leq \frac{1}{n^3}$, by Weiertsrass M-test, the series $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}sin(nx)$converges absolutely and uniformly on $\mathbb{R}$.

Second, for all $n\in \mathbb{N}$, $\frac{1}{n^3}sin(nx)$ is continuous on $[0,2\pi]$. Therefore, the series itself is continuous(also uniformly continuous) on $[0,2\pi]$ due to uniformly convergence.

Last, it's clear that for all $n$, $2\pi$ is a period for $\frac{1}{n^3}sin(nx)$. Hence $2\pi$ is also a period for the series $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}sin(nx)$ (you can prove it by the definition of limit, which is easy). A continuous and periodic function on $\mathbb{R}$ is uniformly continuous.

Proving $\sum^\infty_{n=1} \frac{1}{n^3} \sin(nx)$ is uniformly continuous

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