$\frac{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)(\sin(n x))}{n}$ converges conditionally for $x$ being non-integer multiple of $\pi$

Question

Summation $$\frac{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)(\sin(n x))}{n}$$ converges conditionally for $x$ being non-integer multiple of $\pi$. I can be easily shown that it converges absolutely for $x$ being integer multiple of $\pi$.

Answer 1

Partial sum of harmonic series is $$0<H_n \leq \ln{n}+1$$ $$\frac{\ln{n}}{n} \rightarrow 0, n\rightarrow\infty$$ and $|\sin{(nx)}|\leq 1$. As a result $$\color{red}{0}\leq\left|\frac{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\cdot(\sin(n x))}{n}\right|=\\ \left| \frac{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)}{n}\right|\cdot \left| \sin{(nx)}\right|\leq \\ \frac{\ln{n}+1}{n}\rightarrow \color{red}{0}, n\rightarrow\infty$$ according to squeeze theorem it converges for any $x\in \mathbb{R}$.