The following question appeared on an old complex analysis exam. (The only tools we have learned are simple series for $\sin$, $\cos$, $\exp$ etc., the residue theorem and some tools from real analysis like the Weierstrass M-test.)
Evaluate $\displaystyle \sum_{n\geqslant 1} \frac{(-1)^n \sin(n\theta)}{n^3}$ for $\theta\in(-\pi,\pi)$.
I first believed this had to be evaluated using the residue theorem, but after seeing this thread I tried to manipulate it by taking the derivative.
Since $\displaystyle \left\vert\frac{(-1)^n\sin(n\theta)}{n^3} \right\vert\leq \frac{1}{n^3}$ and $\sum_{n\geq 1}1/n^3<\infty$, it is uniformly convergent for all $\theta$ by the Weierstrass M-test.
We can rewrite the sum as $\operatorname{Im}\sum_{n\geq 1} \frac{(-e^{i\theta})^n}{n^3}$. I tried differentiating the sum, to make use of the closed form $\sum_{n\geq 1}z^n/n=-\log(1-z)$ for $|z|<1$, but this only becomes a big mess.
What is the right way to evaluate this sum using complex analysis?