$\sum_{k=1}^n\frac{\sin kx}{k^\alpha} >0\quad\text{for all}\ n=1,2,3,\ldots\ \text{and}\ 0<x<\pi, \text{and}\ \alpha \ge 1$

Question

The Fejer-Jackson inequality as follows: $$\sum_{k=1}^n\frac{\sin kx}k>0\quad\text{for all}\ n=1,2,3,\ldots\ \text{and}\ 0<x<\pi.$$

I conjecture that the inequality as follows holds:

$$\sum_{k=1}^n\frac{\sin kx}{k^\alpha} >0\quad\text{for all}\ n=1,2,3,\ldots\ \text{and}\ 0<x<\pi, \text{and}\ \alpha \ge 1$$

How to prove this inequality? Can You give a comment or a proof or a reference?

Answer 1

Comment by Cherng-tiao Perng converted to an answer: It appears that Theorem A of this paper solves your problem.