Uniform convergence of $\sum -\frac{\sin(nx)}{\log(n)^2}$ over a closed and bounded subinterval of $[0,2 \pi]$

Question

As mentioned in the title, I want to study the uniform convergence of the following series. $\sum -\frac{\sin(nx)}{\log(n)^2}$ for $x\in[a,b]$ where $0<a<b<2\pi$

I try to apply $M$-test but I am get $\left|\frac{\sin(nx)}{\log(n)^2}\right|\leq\frac{1}{\log(n)^2}$

As $\sum \frac{1}{\log(n)^2}$ diverge, I am going nowhere.

Could anyone help me or giving me some hints?

Thank you very much for reading my passage.

summation by parts using that $\Sigma {\sin(nx)}=O(\frac{2\pi}{a}), 0<a<x<b < 2\pi$ — Conrad, Apr 24 '19 at 13:58

score 1 · Answer 1 · answered Apr 24 '19 at 13:57

You can apply Dirichlet's test for uniform convergence:

$\left(\sum_{n=1}^N\sin(nx)\right)_{N\in\mathbb N}$ is uniformly bounded;
the sequence $\left(\frac1{\log^2n}\right)_{n\in\mathbb N\setminus\{1\}}$ is monotonic;
$\lim_{n\to\infty}\frac1{\log^2n}=0$.

Uniform convergence of $\sum -\frac{\sin(nx)}{\log(n)^2}$ over a closed and bounded subinterval of $[0,2 \pi]$

1 Answers1