Examining if
$$\sum_{n=1}^\infty \frac{\sin n\theta}{n}$$
converges for\footnote{Clearly this covers the useful
cases, as if $\theta$ is a multiple of $2\pi$ or is not real, this
clearly won't converge.}
$\theta \in (0,2\pi)$
is rather delicate.

Spivak does it using Dirichlet's test;
another (equivalent???) approach
(following Alf van der Poorten, ``Life on the Edge'', Monthly, Nov.~2002)
is to look at the power series for
$$-\log (1-z) = \sum_{n=1}{\infty} \frac{z^n}{n}$$
and plugging in $z=e^{i\theta}=\cos \theta + i\sin \theta$
(plus some manipulations) yields
$$\sum_{n=1}^\infty \frac{\cos n\theta}{n} = -\log\left(2\sin \frac{\theta}{2}\right)$$
$$\sum_{n=1}^\infty \frac{\sin n\theta}{n} =
\frac{\pi}{2}-\frac{\theta}{2}$$
