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!
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 $x \in 2\pi \Bbb{Z}$.