
> 
> 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:


[$\sum\_{k=1}^n\frac{\sin kx}{k^\alpha} >0\quad\text{for all}\ n=1,2,3,\ldots\ \text{and}\ 0<x<\pi, \text{and}\ \alpha \ge 1$](https://mathoverflow.net/questions/303759/sum-k-1n-frac-sin-kxk-alpha-0-quad-textfor-all-n-1-2-3-ldots-te?noredirect=1&lq=1)



$\sum_{k=1}^n(-1)^k\frac{\sin kx}{k^{\alpha}} < 0 ?$ with $\alpha \ge 1$ and $n=1, 2,\cdots$
Plug in $\pi-x$ in the already proven inequality, i.e. the one without the $(-1)^k$ one use that
$$
sin(k(\pi-x))=sin(k\pi-kx)=-(-1)^ksin(kx)
$$


