$\sum_{n \neq 0} {\dfrac{\sin(n\vartheta)}{n}}$ does not converge absolutely, when $0 < \vartheta < \pi$.

Question

I want to show that the series $\sum_{n \neq 0} {\frac{\sin(n\theta)}{n}}$ does not converge absolutely when $0 < \theta < \pi$. When $\theta$ is of the form $\frac{a}{b}\pi$ with $a,b \in \mathbb{Z}$, it is easy to show this is true, since in this case the series is bounded below by a multiple of the harmonic series. However, I can't seem to show that the series diverges for all values $0 < \theta < \pi$. It doesn't seem that a continuity argument will work.

Answer 1

Observe that $$ \sum_{k=1}^n \frac{\lvert \sin(k\vartheta)\rvert}{k}\ge \sum_{k=1}^n \frac{\sin^2(k\vartheta)}{k} =\sum_{k=1}^n \frac{1-\cos(2k\vartheta)}{2k}\\=\frac{1}{2}\sum_{k=1}^n\frac{1}{k}-\frac{1}{2}\sum_{k=1}^n\frac{\cos(2k\vartheta)}{k}\to \infty, $$ as $n\to\infty$, since the sequence $$ s_n=\sum_{k=1}^n\frac{\cos(2k\vartheta)}{k} $$ converges, if $\vartheta\ne m\pi$. (Try for example, Abel Test.)