absolutely convergent & conditionally convergent

Question

Prove that

$$\sum_{n=1}^{\infty}\left(\sum_{m=1}^{n}\frac{{{1}}}{{{m}}}\right)\frac{{{\sin(nx)}}}{{{n}}} $$

for $x = {k\pi}$ , $k\in \mathbb{Z} $ is absolutely convergent.

&

for $x \not= {k\pi}$ , $k\in \mathbb{Z}$ is conditionally convergent.

Answer 1

If $x\in\pi\mathbb{Z}$ we are just summing zeroes, so there is little to prove. If $x\not\in\pi\mathbb{Z}$, from: $$\sum_{n=1}^{N}\sin(nx)=\frac{\sin\frac{Nx}{2}\sin\frac{(N+1)x}{2}}{\sin\frac{x}{2}}\in\left[-\frac{1}{2}\tan\frac{x}{4},\frac{1}{2}\cot\frac{x}{4}\right]$$ and the fact that $\frac{H_n}{n}$ is eventually decreasing and converging to zero, we get that the original series is conditionally convergent by Dirichlet's test or just summation by parts. In such a case, however, the series is not absolutely convergent since, given that $\pi a$ is the closest element of $\pi\mathbb{Z}$ to $x$ and $d=\left|a -\frac{x}{\pi}\right|\in\left(0,\frac{1}{2}\right]$, the asymptotic density of the integers $m$ such that $|\sin(mx)|\geq\frac{1}{2}$ can be lower-bounded in terms of $d$.

For $x\not\in\pi\mathbb{Z}$, we have: $$ f(x) = \sum_{n\geq 1}\frac{H_n}{n}\sin(nx)=\text{Im}\left(\frac{1}{2}\log^2(1-e^{ix})+\operatorname{Li}_2(e^{ix})\right).$$

^{$H_n$ stands for the $n$-th harmonic number: $H_n=\sum_{k=1}^{n}\frac{1}{k}$.}