Sum of series $\sin \theta+\sin 2 \theta+\sin 3\theta+\dots$ I need to prove sum of series: $$\sin \theta+\sin 2 \theta+\sin 3\theta+\dots=\sum_{n=1}^\infty \sin n\theta$$
by using in the first place the complex numbers.

$$\sum_1^{\infty} \sin (n\theta) = \sum_1^{\infty} \frac{1}{2i} (e^{i\theta} - e^{-i\theta}) = 
\frac{1}{2i} \left( \sum_1^{\infty}e^{i\theta}  - \sum_1^{\infty}e^{-i\theta} \right) \\ = \frac{1}{2i} \left( \frac{e^{i\theta}}{1-e^{i\theta}} - \frac{e^{-i\theta}}{1-e^{-i\theta}}  \right) 
$$
$$
= \frac{\frac{1}{2i} (e^{i\theta}-e^{-i\theta})}{2-(e^{i\theta}+e^{-i\theta})} \\
= \frac{\sin\theta}{2(1-\cos\theta)}
$$
