The following Dirichlet series converges for $\Re(s)\in (1,\infty)$ (constant $k\in\mathbb{R}$), which is evident since sine function in numerator is in interval $[-1,1]$. $$f_1(s)=\sum _{n=1}^{\infty } \frac{\sin \left(\frac{2 \pi k}{n}\right)}{n^s}$$
How do we make analytic continuation of it for $\Re(s)<1$?
I saw a quite complicated method how to make analytic continuation of $$f_2(s)=\sum _{n=1}^{\infty } \frac{\sin \left(2 \pi k n\right)}{n^s}=\frac{1}{2} i \left(\text{Li}_s\left(e^{-2 i k \pi }\right)-\text{Li}_s\left(e^{2 i k \pi }\right)\right)$$ using generalized polylogarithm function.
The second series differs from the first series by having $n$ in numerator instead of denominator inside sine function. And this fact seems to prevent using the same method.