Show that $\sum_{n=0}^\infty \frac{\sin {nx}}{a^n} = \frac{a \sin{x}}{1 + a^{2} - 2a \cos{x}}$

Question

Show that $$\sum_{n=0}^\infty \frac{\sin {nx}}{a^n} = \frac{a \sin{x}}{1 + a^{2} - 2a \cos{x}}$$

I've been trying to use the geometric series rule for $\sum_{n=0}^\infty x^{-n} = \frac{x}{x -1}$ as well as Euler for the $\sin(nx)$, but I just can't seem to get the series to reduce to the fraction on the right. Could you help?

Please, include the problem in the body of the question too. — Pedro, Oct 07 '12 at 20:20

score 17 · Answer 1 · answered Oct 07 '12 at 18:21

17

Look at the imaginary part of $$\sum_{n=0}^{\infty} \dfrac{\exp(inx)}{a^n} = \sum_{n=0}^{\infty} \left(\dfrac{\exp(ix)}{a} \right)^n = \dfrac{a}{a-\exp(ix)}$$

answered Oct 07 '12 at 18:21

hehe, nice (+1) – user 1591719 Oct 07 '12 at 18:50
Thanks Marvis, that was super helpful! That's a really clever maneuver taking the n out of the whole phrase. – Jason Oct 07 '12 at 19:18

Show that $\sum_{n=0}^\infty \frac{\sin {nx}}{a^n} = \frac{a \sin{x}}{1 + a^{2} - 2a \cos{x}}$

1 Answers1