Convergence of the series $\frac{1} {n} \sin(nx) $

Question

I've been trying to work out the solution to this problem. Consider $\forall x \in \mathbb R $ the series $$\sum_{n=1}^{+\infty} \frac{1} {n} \sin(nx) $$ Find the values of $x$ for which the series converge (pointwise, uniformly or absolutely).

My attempt:

It is obvious that for every $x \in \mathbb R $, we have $ \vert \sin(nx) \vert \le 1$, and if we take the points $x=\frac{\pi} {2n}$ our series is exactly the harmonic series and it doesn't pass the Weierstrass M-test. So our series is not absolutely and uniformly convergent (at one time).

Then if we take the points $x=k\pi, k \in \mathbb Z$, our series converges to 0, very easily.

For the other values of $x$, I tried using other common tests, such as ratio and root test (I get the inconclusive 1), and figuring out useful bounds for the sequence, but didn't succeed. I suspect it converges, but I don't know how to go on.

Am I mistaken? Are there any other tests?

alternatively to remember some specific criterion you can use summation by parts to solve many questions about convergence or divergence of series. Indeed Dirichlet's criterion is just an application of summation by parts — Masacroso, Oct 29 '20 at 13:32
It also converges, as an alternating series if $x\equiv\frac\pi 2\mod\pi$. — Bernard, Oct 29 '20 at 13:36
What constant I can use to bound the absolute value of the partial sums of $sin(nx)$? — Adriano Banchieri, Oct 29 '20 at 13:43
The constant depends on $x$. So the bound is a function of $x$ (working for all upper limits $k$ of the partial sums) rather than a constant. — Daniel Fischer, Oct 29 '20 at 14:59
I get it. Do you think it might be useful to work out an explicit expression for it? — Adriano Banchieri, Oct 29 '20 at 18:19

Convergence of the series $\frac{1} {n} \sin(nx) $

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