$\sum_{k=1}^n(-1)^k\frac{\sin kx}{k^{\alpha}} < 0 ?$ with $\alpha \ge 1$ and $n=1, 2,\cdots$

Question

Could You give a poof, comment or reference for the inequality as follows:

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

• See also:

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

Answer 1

Plug in $\pi-x$ in the already proven inequality, i.e. the one without the $(-1)^k$ one use that $$ sin(k(\pi-x))=sin(k\pi-kx)=-(-1)^ksin(kx) $$