Q:

$\sum\limits_{n=1}^∞ \frac{|\sin(n\theta)|}{n}$ diverges?

Let $\theta$ be any fixed constant in $(0,\pi/2)$. Then the seires $$\sum_{n=1}^∞ \frac{|\sin(n\theta)|}{n}$$ diverges.

I don't have any idea how to prove it.
Maybe equidistribution theorem is related?

A:

Hint The standard arguments on the density of $n \frac{\theta}{2 \pi}$ modulo 1 can be changed to show that for each $\theta \in (0 , \frac{\pi}{2})$ there exists some $M$ so that among any $M$ consecutive numbers $n+1, n+2,.., n+M$ you can find some $m$ so that 
$$|\sin(m \theta)| \geq \frac{1}{2}$$
Conclude from here that your series is larger than 
$$\sum_{k=1}^\infty \frac{1}{2} \frac{1}{kM}$$

A:

Note that
$$|\sin (nt)| \ge \sin^2 (nt) = \frac{1-\cos (2nt)}{2}.$$
If we show
$$\tag 1 \sum_{n=1}^{\infty}\left( \frac{1}{2n}- \frac{\cos (2nt)}{2n}\right)$$
diverges, we'll be done. Now $(1)$ is the sum of two series, the first of which diverges as you know, the second of which converges by Dirichlet's test. Hence $(1)$ diverges, which implies the original series diverges.

