How to prove the series converge for every $\theta$? How to prove the following series converge for every $\theta$, any suggestions are welcome?

$$ \sum_{n=1}^{\infty}\frac{(-1)^n\sin(n\theta)}{n}.$$

 A: Your series is just the Fourier sine series of the (sawtooth wave) periodic continuation of the function $f(\theta)=-\frac{\theta}{2}$ defined over $[-\pi,\pi]$, hence:
$$\sum_{n=1}^{+\infty}\frac{(-1)^n \sin(n\theta)}{n}=\left\{\begin{array}{rcl}-\frac{1}{2}(\theta\operatorname{mod} 2\pi)&\text{if}&\theta\not\in\pi\mathbb{Z},\\ 0&\text{if}&n\in\pi\mathbb{Z},\end{array}\right.$$
where $x\to x\operatorname{mod} 2\pi$ is intended to map any real number into $[-\pi,\pi]$. This just follows from:
$$-\frac{1}{\pi}\int_{-\pi}^{+\pi}\frac{x}{2}\sin(nx)\,dx = \frac{(-1)^n}{n}$$
that can be checked through integration by parts.

As an alternative, you can prove convergence by noticing that:
$$\sum_{n=1}^N \frac{(-1)^n\sin(n\theta)}{n}=\sum_{n=1}^N \frac{\cos(\pi n)\sin(n\theta)}{n}=\frac{1}{2}\left(\sum_{n=1}^N \frac{\sin((\theta+\pi)n)}{n}+\sum_{n=1}^N \frac{\sin((\theta-\pi)n)}{n}\right)$$
and since the partial sums of $\sin(\psi n)$, for any $\psi\not\in\pi\mathbb{Z}$, are bounded by $\frac{1}{|\sin(\psi/2)|}$, the conditions for Dirichlet's test are met.
A: Hint:  You can use Dirichlet's test.
Added: You need to bound partial sums 

$$ \sum_{i}^{n} (-1)^i\sin(i\theta) = {\frac { \left( -1 \right) ^{n}\sin \left( \theta\,n \right) + \left( 
-1 \right) ^{n}\sin \left( \theta\,n+\theta \right) -\sin \left( 
\theta \right) }{2+2\,\cos \left( \theta \right) }}$$

which is easy to do.
