Does anyone know of a way to simplify this sum?

$$S(n)=\sum_{j=1}^{\rho(n)}\sum_{k=1}^\infty\frac{\sin[2\pi k n 2^{-j}]-\sin[2\pi k (n-1) 2^{-j}]}{k}$$

where $\rho(n)=[\log_2(n)]$ (and $[x]$ denotes the greatest integer less than $x$).

Note: This question is a follow-up to a previous question I asked: Greatest power of two dividing an integer

EDIT: After following all the given suggestions, I found that for integer $n$,

$$\frac{S(n)}{\pi}=2^{-\rho(n)}-1+\frac{1}{1+(-1)^n}\sum_{j=1}^{\rho(n)}\left[\frac{n}{2^j}\right]-\left[\frac{n-1}{2^j}\right].$$

This is pretty much what I started with in my previous post, so if anyone knows of a way to take this sum, please let me know. Anyway, I will leave this result here in case anyone ever comes across $S(n)$ in some other context. Thanks to everyone who helped.