Is $\sum_{n=1}^{\infty}{\frac{\sin(nx)}n}$ continuous? Considering the infinite series $\sum_{n=1}^{\infty}{\frac{\sin(nx)}n}$ , I can show that  it is not convergent uniformly by Cauchy's criterion and that it is convergent for every $x$ by Dirichlet's test.  But I don't know how to judge whether it is continuous.
Could you tell me the answer and why? Thank you in advance!
 A: To inspect the discontinuity of the summation, let's calculate the sum. By the Abel's theorem,
$$ f(x) := \sum_{n=1}^{\infty} \frac{\sin nx}{n} = \lim_{s\to 0^{+}} \sum_{n=1}^{\infty} \frac{\sin nx}{n} e^{-ns}. $$
By utilizing Taylor expansion of the logarithm,
\begin{align*}
\sum_{n=1}^{\infty} \frac{\sin nx}{n} e^{-ns}
&= \Im \sum_{n=1}^{\infty} \frac{e^{n(ix-s)}}{n}
 = - \Im \log (1 - e^{ix-s}) \\
&= -\Im \log (1 - e^{-s}\cos x - ie^{-s}\sin x) \\
&= \arctan \left(\frac{e^{-s}\sin x}{1 - e^{-s}\cos x}\right).
\end{align*}
Thus taking $s \to 0^{+},$
$$ f(x) = \arctan \left(\frac{\sin x}{1 - \cos x}\right) = \arctan \left(\cot \frac{x}{2}\right) = \arctan \left(\tan \frac{\pi-x}{2}\right). $$
Therefore
$$ f(x) = \begin{cases}
\frac{\pi - x}{2} & x \in (0, 2\pi),\\
0 & x = 0, \\
f(x+2\pi), & x \in \Bbb{R}.
\end{cases} $$
This shows a clear-cut jump discontinuity at each $x \in 2\pi \Bbb{Z}$.
A: Another way to prove that $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ is discontinuous at the origin (and so at every element of $2\pi\mathbb{Z}$).
If $f(x)$ is a bounded function and $\lim_{x\to 0^+}f(x)$ does exist, it equals $\lim_{m\to +\infty}\int_{0}^{+\infty} f(x)m e^{-mx}\,dx $ (approximation of the identity). Due to
$$ \int_{0}^{+\infty}\sin(nx)m e^{-mx}\,dx = \frac{m n}{m^2+n^2} $$
we have
$$ \lim_{x\to 0^+}\sum_{n\geq 1}\frac{\sin(nx)}{n} = \lim_{m\to +\infty}\sum_{n\geq 1}\frac{m}{m^2+n^2} $$
and by Riemann sums the RHS of the last line equals $\int_{0}^{+\infty}\frac{dx}{x^2+1}=\frac{\pi}{2}\neq 0$.
