A trigonometric series Let $\alpha$ be a real number. I'm asked to discuss the convergence of the series 
$$
\sum_{k=1}^{\infty} \frac{\sin{(kx)}}{k^\alpha}
$$
where $x \in [0,2\pi]$. 
Well, I show you what I've done:


*

*if $\alpha \le 0$ the series cannot converge (its general term does not converge to $0$ when $k \to +\infty$) unless $x=k\pi$ for $k=0,1,2$. In other words, if $\alpha \le 0$ there is pointwise convergence only in $x=0,\pi,2\pi$.

*if $\alpha \gt 1$, I can use the Weierstrass M-test to conclude that the series is uniformly convergent hence pointwise convergent for every $x \in [0,2\pi]$. Moreover the sum is a continuous function in $[0,2\pi]$. 
Would you please help me in studying what happens for $\alpha \in (0,1]$? Are there any useful criteria that I can use?
Does the series converge? And what kind of convergence is there? In case of non uniform but pointwise convergence, is the limit function continuous?
Thanks.
 A: Note that $$\sum_{k=1}^{N} \sin(kx) = \dfrac{\sin(Nx/2)}{\sin(x/2)} \sin \left( \left(\dfrac{N+1}2 \right)x\right)$$ Hence, for each given $x$, the sum is bounded by $\dfrac1{\sin(x/2)}$.
Hence by generalized alternating series test (also known as Dirichlet's test) the sum converges.
A: There is an easier proof that $f_\alpha$ is discontinuous at 0.
Let $x=\pi/n$ for some even $n$. Then for $1\le i\le n$, group terms $a_{2nk+i}$ and $a_{2nk+i+n}$ together:
$$\frac{\sin (2nk+i)x}{(2nk+i)^\alpha}+\frac{\sin (2nk+i+n)x}{(2nk+i+n)^\alpha}\ge\frac{n\alpha}{(2nk+i)(2nk+i+n)^\alpha}\sin \frac{i\pi}{n} \ge 0$$
We need only use $k=0$ and $i\le n/2$ for the lower bound:
$$\begin{align}
f_\alpha(x)\ge&\alpha \sum_{i=1}^{n/2} \frac{n}{i(i+n)^\alpha}\sin \frac{i\pi}{n}\\
\ge&\alpha\sum_{i=1}^{n/2} \frac{2}{(i+n)^\alpha}\qquad{\text{($\sin t\ge 2/\pi\cdot t$)}}\\
\ge&2\alpha\log\frac{3n/2+1}{n+1}\\
\rightarrow&2\alpha\log 3/2
\end{align}$$
So $f_\alpha$ is bounded from below by a positive number as $x\rightarrow 0$, for all $\alpha\in(0,1]$. Because $f_\alpha(0)=0$, the convergence cannot be uniform around 0.
