Does $\sum_{k=1}^\infty(1+\frac{1}{2}+ \dots +\frac{1}{k})\frac{\sin kx}{k}$ converge absolutely? For $x\neq k\pi,k\in\mathbb{Z}$,does the following series converge absolutely?
$$\sum_{k=1}^\infty(1+\frac{1}{2}+ \dots +\frac{1}{k})\frac{\sin kx}{k}$$

Does it converge absolutely even without the $\left(1+\cdots+\frac{1}{k}\right)$? I doubt it. For most fixed $x$, a nearly arithmetic progression of $k$-values will leave $\left|\sin kx\right|$ very close to $1$, and so a subseries would involve a series with similar features to the Harmonic series. Tossing in the $\left(1+\cdots+\frac{1}{k}\right)$ just makes it less able to converge.
(Whether it converges conditionally is a different question.)
