I've found the limit of a similar expression in the literature (Abramowitz & Stegun), which uses Bernoulli polynomials:
$\sum_{j=1}^{\infty} \frac{1}{j^{2k+1}}\sin{2\pi j x}=\frac{-(-1)^k(2\pi)^{2k+1}}{2(2k+1)}B_{2k+1}(x)$
My question is, how about the below, does it also have a known limit? (The aforementioned reference doesn't have it.)
$\sum_{j=1}^{\infty} \frac{1}{j^{2k}}\sin{2\pi j x}$
I would guess not, but unlikely. Reason is cause this is similar to the non-existence of closed-forms for say $\zeta(2k+1)$ as opposed to $\zeta(2k)$, but need to confirm. (Yes, searching the literature for the novelty of things is one of the most challenging tasks.)