prove that $\sum_{n = 1}^{\infty} \frac{|\sin(nx)|}{n}$ is divergent

Question

I tried much to show that the following series is divergent for $x \in (0,\pi)$

$\sum_{n = 1}^{\infty} \frac{|\sin(nx)|}{n}$

My opinion was to use the comparison test with series $\frac{1}{n}$ but I could not achieve something good.

Thank for your time.

Answer 1

Proof using Dirichlet's test: Firstly I'll introduce the test:

Given a sequence $\{a_n\}_{1}^{\infty}$ and a sequence $\{b_n\}_{1}^{\infty}$ satisfying:

• $\{a_n\}_{1}^{\infty}$ is monotonic

• $\lim_{n \to \infty}{a_n} = 0$

• $ \left| \sum_{n=1}^{N} b_n \right| \leq M $ for every positive integer $N$

The series $ \sum_{n=1}^{\infty} a_n \cdot b_n $ converges.

Let us define $ a_n = \frac{1}{n} $ and $ b_n = \left| sin(n\cdot x) \right| $.

We know that $a_n$ is monotonic and converges to 0.

All we have left to prove is that $ \left| \sum_{n=1}^{N} b_n \right| = \left| \sum_{n=1}^{N} \left| sin(n\cdot x) \right| \right| = \sum_{n=1}^{N} \left| sin(n\cdot x) \right| \leq M $ for every positive integer $N$

We know that $\left| sin(n \cdot x)\right| \leq 1$ and so $\sum_{n=1}^{N} \left| sin(n\cdot x) \right| \leq \sum_{n=1}^{N} 1 = N $

And so for every positive integer $N$:

$ \left| \sum_{n=1}^{N} b_n \right| \leq N $

According to Dirichlet's test, we get that:

$ \sum_{n=1}^{\infty} a_n \cdot b_n = \sum_{n=1}^{\infty} \frac{ \left| sin(n\cdot x) \right| }{n} $ converges.

$$\blacksquare$$