This is from an exercise on 16.2.4 of Mathematical Analysis by Zorich:
Investigate the nature of the convergence on the sets $E\in \mathbb R$ for different values
of the real parameter $\alpha$ in the following series:
a)$\sum_{n=1}^{\infty}\frac{\cos nx}{n^{\alpha}}$ and b) $\sum_{n=1}^{\infty}\frac{\sin nx}{n^{\alpha}}$
The Abel-Dirichlet test deduce that when $0<\alpha\leq 1$, $\frac{\sin nx}{n^\alpha}$ converges in $\mathbb R$, because it converges uniformly in the set E such that $\inf\limits_{x\in E}\sin |\frac x2|>0 $, and converges on the point $x=k\pi,k\in \mathbb Z$. Hence we also know it converges uniformly in the any compact set $K$ s.t. $k\pi\notin K\; (\forall k\in \mathbb Z)$.
But, what if a compact set $K$,or the closure of a bounded set $E$ contains some points of the form $2k\pi$? For example, does $\frac{\sin nx}{n^\alpha}$ converge uniformly in $(0,\pi)$ or $[0,\pi]$?
Feel free to give some hints or point out any mistakes above.
Best Answer
Convergence is not uniform on intervals where $2k\pi$ is a limit point when $0 < \alpha \leqslant 1$.
Consider for example $(0,\pi)$. For any $m \in \mathbb{N},$ let $x_m = \pi/(4m)$. With $m < n \leqslant 2m$, we have $\pi/4 < nx_m \leqslant \pi/2$ and $1/ \sqrt{2} < \sin n x_m \leqslant 1 $.
Hence,
$$\left|\sum_{n = m+1}^{2m} \frac{\sin nx_m}{n^\alpha}\right| > \frac{1}{\sqrt{2}}\sum_{n=m+1}^{2m}\frac{1}{n^\alpha}> \frac{1}{ \sqrt{2}} \cdot (2m- m)\cdot \frac{1}{(2m)^\alpha}= \frac{m^{1-\alpha}}{2^{\alpha}\sqrt{2}}\\ \underset{m \to \infty}\longrightarrow \begin{cases}+\infty,&0 < \alpha <1\\ \frac{1}{2\sqrt{2}}, & \alpha =1 \end{cases} \neq 0,$$
and the series fails to converge uniformly by violation of the uniform Cauchy criterion. The argument can be modified to prove non-uniform convergence on intervals where $2k\pi$ is a limit point with $k \neq 0$.