$\sum\limits_{n=1}^{\infty}\frac{\sin(nx)}{n^{\alpha}}$ does not converge uniformly for $\alpha \in (0,1]$ 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
 A: Alternative solution:
First, proving the convergence of the $\sum_{n=1}^{\infty}\frac{\sin{nx}}{n^{\alpha}}$ for $0<\alpha\le 1$
Since $\sin(nx)$ is periodic, just consider $x \in (-\pi,\pi)$
Use Dirichlet's test:
{$a_{n}$} is a sequence of real numbers and {$b_{n}$} a sequence of complex numbers satisfying


*

*$a_{n+1}\le a_{n}$

*$\lim_{n \to \infty} a_{n}=0$

*$\left| {\sum_{n=1}^{N}b_{n}} \right| \le M$ for every positive integer $N$
where $M$ is some constant, then the series
$$\sum_{n=1}^{\infty} a_{n}b_{n}$$
converges.
details in this link
Let {$\frac{1}{n^{\alpha}}$} be {$a_{n}$} and let {$\sin(nx)$} be {${b_{n}}$}
It's easy to see that the first 2 points satisfied. For $3.$ :
$$\sum_{n=1}^{N}\sin(nx)=\bar {D}_{N}(x)$$
 For 
$$\bar {D}_{N}(x)=\sum_{|n|\le N}\operatorname{sign}(x)e^{inx}/i \quad \text{where } \operatorname{sign}(x)=\begin{cases} 1 & \text{if } n>0 \\ 0 & \text{if } n=0 \\ -1\ & \text{if } n<0.\end{cases}$$
and
$$\bar {D}_{N}(x)=\frac{\cos(x/2)-\cos((N+1/2)x)}{\sin(x/2)},$$
so
$$\left| \sum_{n=1}^{N}\sin(nx)\right|\le \frac{2}{\left| \sin(x/2) \right|}$$
For the continuity of $\sum_{n=1}^{\infty}\sin(nx)/n^{\alpha}$,check this link
