# In what set does $\sum_{n=1}^{\infty}\frac{\sin nx}{n^{\alpha}}$ converge uniformly

calculusconvergence-divergencereal-analysisuniform-convergence

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.

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 $$1/ \sqrt{2} < \sin n x_m \leqslant 1$$.
$$\left|\sum_{n = m+1}^{2m} \frac{\sin nx_m}{n^\alpha}\right| > \frac{1}{\sqrt{2}}\sum_{n=m+1}^{2m}\frac{1}{n^\alpha}> \frac{1}{ \sqrt{2}} \cdot (2m- m)\cdot \frac{1}{(2m)^\alpha}= \frac{m^{1-\alpha}}{2^{\alpha}\sqrt{2}}\\ \underset{m \to \infty}\longrightarrow \begin{cases}+\infty,&0 < \alpha <1\\ \frac{1}{2\sqrt{2}}, & \alpha =1 \end{cases} \neq 0,$$
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 $$k \neq 0$$.