The question is, why $\sum\limits_{n=1}^{\infty}\frac{\sin(nx)}{n^{\alpha}}$ does not converge uniformly for $\alpha \in (0,1]$. I have a solution, but I dont understand the estimates, sorry=(. Consider $x_k=\frac{\pi}{2^k}$, it is $\sum\limits_{n=1}^{2^k}\frac{\sin\left(n\frac{\pi}{2^k}\right)}{n^{\alpha}}\ge \sum\limits_{n=2^{k-2}}^{3\cdot2^{k-2}}\frac{\sin\left(n\frac{\pi}{2^k}\right)}{n^{\alpha}}\ge \frac{1}{\sqrt{2}}\sum\limits_{n=2^{k-2}}^{3\cdot2^{k-2}}\frac{1}{n^{\alpha}}\ge \frac{1}{\sqrt{2}}2^{k-1}\frac{1}{(3\cdot2^{k-2})^{\alpha}} $. Could you explain me, why
(1)$\sum\limits_{n=1}^{2^k}\frac{\sin\left(n\frac{\pi}{2^k}\right)}{n^{\alpha}}\ge \sum\limits_{n=2^{k-2}}^{3\cdot2^{k-2}}\frac{\sin\left(n\frac{\pi}{2^k}\right)}{n^{\alpha}}$
(2)$\frac{1}{\sqrt{2}}\sum\limits_{n=2^{k-2}}^{3\cdot2^{k-2}}\frac{1}{n^{\alpha}}\ge \frac{1}{\sqrt{2}}2^{k-1}\frac{1}{(3\cdot2^{k-2})^{\alpha}} $
is true?
Or could you give me an alternative solution? Regards