Series $ \sum\limits_{n=1}^{\infty}\frac{1}{n}e^{2 \pi i n x} $ Prove that the sum
  \begin{equation}
\sum\limits_{n=1}^{\infty}\frac{1}{n}e^{2 \pi i n x}
\end{equation}
  converges for any $ x \notin \mathbb{Z} $.




By definition, the initial series converges if and only if $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\cos{(2 \pi n x)} $ and $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\sin{(2 \pi n x)} $ both converge. Neither Dirichlet's nor Abel's test works. Integral test can't be applied because $ \frac{1}{t}\cos{(2 \pi x t)} $ is not monotonic function of $ t $. This problem slightly reminds me of how Equidistributional theorem is proved, but most probably it has nothing to do with it.
Hint We have for $x\neq 2\pi m$; $$\sum_{k=1}^ne^{ikx}=\frac{\sin(nx/2)}{\sin(x/2)}e^{i(n+1)x/2}$$

Thus Dirichlet's criterion works all right. This says the following:


  Let $\langle a_n\rangle$ be a sequence of complex numbers with bounded partial sums, and let $b_n$ be a sequence of real numbers decreasing monotonically to zero. Then $$\sum a_nb_n$$ converges.
Let $a_n(x)=\sum_{k=1}^n {\rm e}^{{\rm i}kx}$; even though  $\lim_{x\rightarrow 0}|a_n(x)|=n$ (and $|a_n(\pi)|=|\sin(n\frac{\pi}{2})|$), for each $x$ in $(0, \pi]$,  $ a_n(x)$ is indeed bounded. In fact, for large fixed $r$, $|a_n(\frac{\pi}{r})|=|\frac{\sin(\frac{n\pi}{2r})}{\sin(\frac{\pi}{2r})}|$ $\approx|\frac{\sin(\frac{ n\pi}{2r})}{\frac{\pi}{2r}}|$ $={\frac{2r}{\pi}}|\sin(\frac{n\pi}{2r})|$ is bounded. Also, the local maxima of $|a_n(x)|$, for $x\in(0, \pi]$, decrease with $x$.