$\sum_{k=1}^{+\infty} \frac{ \mid sin (kx) \mid}{k}$

Question

What is the nature of : $$\sum_{k=1}^{+\infty} \frac{ \mid sin (kx) \mid}{k}$$

We know that :

1. $\sum_{k=1}^{+\infty} \frac{ sin (kx)}{k}$ converges with Dirichlet test proof
2. $\sum_{k=1}^{+\infty} \frac{ sin ^2(kx)}{k}$ diverges proof
3. $\sum_{k=1}^{+\infty} \frac{ sin ^2(kx)}{k^2}$ converges

Answer 1

Actually, we have $\sin(kx)^2\le|\sin(kx)|$ (Why ?)

So $\sum_{k=1}^{+\infty} \frac{ \mid \sin (kx) \mid}{k} \ge \sum_{k=1}^{+\infty} \frac{ \sin (kx)^2}{k}$ which, I think, concludes on the divergence.