Does the sum $\sum\limits^{\infty}_{k=1} \frac{\sin(kx)}{k^{\alpha}}$ converge for $\alpha > \frac{1}{2}$ and $x \in [0,2 \pi]$? Does the series
$$
\sum^{\infty}_{k=1} \frac{\sin(kx)}{k^{\alpha}},
$$ 
converge for all $\alpha > \frac{1}{2}$  and for all $x \in [0,2 \pi]$? 
It is obvious that it does when $\alpha > 1$, but I have no idea how to deal with the case
$$
\frac{1}{2} < \alpha \le 1. 
$$ 
I already appreciate your hints/ideas.
 A: The set $\{ e^{inx} \, | \, n \in \mathbb Z \}$ forms an Hilbert basis of $L^2([0,2\pi])$ which means that for a function $f \in L^2([0,2\pi])$, we have
$$
f = \sum_{n \in \mathbb Z} \langle f, e_n \rangle e_n
$$
with 
$$
\langle f,e_n\rangle = \frac{n/|n|}{2i n^{\alpha}}
$$
if $n \neq 0$ and $0$ otherwise, if and only if
$$
\sum_{n \in \mathbb Z} |\langle f,e_n \rangle|^2 < \infty.
$$
Since $\alpha > \frac 12$, in our situation this is the case, therefore $\sum_{n \in \mathbb Z} \frac{\sin(nx)}{n^{\alpha}}$ is well-defined and converges pointwise almost everywhere to some function $f(x)$ which is in $L^2([0,2\pi])$. 
Hope that helps,
A: Actually, this sum converges for every $\alpha>0$.
Step I. For every $x\in\mathbb R$, the sequence $s_n=\sum_{k=1}^n\sin kx$ is bounded. 
Indeed, if $x=m\pi$, then $s_n=0$. If  $x\ne m\pi$, then $\sin(x/2)\ne 0$, and
$$
s_n=\sum_{k=1}^n\sin kx=\mathrm{Im}\left(\mathrm{e}^{xi}+\mathrm{e}^{2xi}+\cdots\mathrm{e}^{nxi}\right)=
\mathrm{Im}\left(\mathrm{e}^{xi}\frac{\mathrm{e}^{nxi}-1}{\mathrm{e}^{xi}-1}\right)
$$
But
$$
\left|\mathrm{e}^{xi}\frac{\mathrm{e}^{nxi}-1}{\mathrm{e}^{xi}-1}\right|\le \frac{2}{|\mathrm{e}^{xi}-1|}=\frac{2}{|\mathrm{e}^{xi/2}-\mathrm{e}^{-xi/2}|}=\frac{1}{|\sin (x/2)|}.
$$
and hence $\lvert s_n\rvert\le \lvert\sin(x/2)\rvert^{-1}$. 
Step II. Use Abel's summation method.
\begin{align}
\sigma_n &=\sum_{k=1}^n\frac{\sin kx}{k^\alpha}=\sum_{k=1}^n\frac{s_k-s_{k-1}}{k^\alpha}
=\sum_{k=1}^n\frac{s_k}{k^\alpha}-\sum_{k=1}^n\frac{s_{k-1}}{k^\alpha} \\
&=\sum_{k=1}^n\frac{s_k}{k^\alpha}-\sum_{k=0}^{n-1}\frac{s_{k}}{{(k+1)}^\alpha}=
\frac{s_n}{n^a}+\sum_{k=1}^{n-1}s_k\left(\frac{1}{k^a}-\frac{1}{(k+1)^a}\right).
\end{align}
But
$$
\frac{1}{k^\alpha}-\frac{1}{(k+1)^\alpha}\le\frac{a}{k^{1+\alpha}},
$$
and hence the series
$$
\sum_{n=1}^{\infty}s_n\left(\frac{1}{n^a}-\frac{1}{(n+1)^a}\right),
$$
converges (indeed absolutely) due to the comparison test.
Note. This series converges conditionally and pointwise. It does not converge absolutely, but it does converge uniformly far from zero, i.e., in any interval
$[\varepsilon,2\pi-\varepsilon]$.
