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.
 A: De Moivre's Theorem states that $e^{ix}=\cos{x} + i \sin{x}$, so $\sin{x} = \Im{(e^{ix})}$ (i.e. the imaginary part of the expression).
So,
$\sin{\theta} + \sin{2\theta} + \sin{3\theta} + \ldots = \Im{(e^{i\theta})} + \Im{(e^{2i\theta})} + \Im{(e^{3i\theta})} + \ldots = \Im{(e^{i\theta} + e^{2i\theta} + e^{3i\theta} + \ldots)}$
Using the fact that taking the imaginary part is a linear operation. And the thing at the end is a geometric series with first term and ratio both equal to $e^{i\theta}$, which should be enough to get you the answer.
A: $$\sum_1^{\infty} \sin (n\theta) = \sum_1^{\infty} \frac{1}{2i} (e^{i\theta} - e^{-i\theta}) = 
\frac{1}{2i} \left( \sum_1^{\infty}e^{i\theta}  - \sum_1^{\infty}e^{-i\theta} \right) \\ = \frac{1}{2i} \left( \frac{e^{i\theta}}{1-e^{i\theta}} - \frac{e^{-i\theta}}{1-e^{-i\theta}}  \right) 
$$
$$
= \frac{\frac{1}{2i} (e^{i\theta}-e^{-i\theta})}{2-(e^{i\theta}+e^{-i\theta})} \\
= \frac{\sin\theta}{2(1-\cos\theta)}
$$
A: We have by elementary trigonometry
$$
2\sin(\frac\theta2)\left(\sin\theta+\sin2 \theta+\dotsb+\sin(n\theta)\right)=2\sin(\frac \theta2)\sin \theta+2\sin(\frac \theta2)\sin 2 \theta+\dotsb+2\sin(\frac \theta2)\sin n\theta=\cos(\frac\theta2)-\cos(\frac{3\theta}2)+\cos(\frac{3\theta}2)-\cos(\frac{5\theta}2)+\dotsb+\cos(\frac{(2n-1)\theta}2)-\cos(\frac{(2n+1)\theta}2)=\cos(\frac\theta2)-\cos(\frac{(2n+1)\theta}2)
$$
But the series is divergent because
$$
\lim_{n\to\infty}\sin(n\theta)\neq0
$$
For $\theta\neq k\pi$.

