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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

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    $\begingroup$ All the terms of $(1)$ are positive, and we are just taking a subset of them. For $(2)$ it is sufficient to notice that $\sin x\geq\frac{1}{\sqrt{2}}$ if $x\in\left[\frac{\pi}{4},\frac{3\pi}{4}\right].$ $\endgroup$ May 12, 2015 at 11:19

1 Answer 1

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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

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

  2. $\lim_{n \to \infty} a_{n}=0$

  3. $\left| {\sum_{n=1}^{N}b_{n}} \right| \le M$ 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

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