Trigonometric Summation $\sum_{r=0}^∞ \frac {\sin(r θ)}{3^r}$

Question

$$\sum_{r=0}^∞ \frac {\sin(r θ)}{3^r} $$

Given: $\sinθ=1/3$

I recently came across this question, and I tried numerous ways of solving it, but I could reach nowhere. My initial approach was to expand uptil the first 3 terms to see if there was a pattern, but I don't think there is one. I also tried converting the question into a telescopic series, but I don't see how that can be done either. Any help would be appreciated!

Answer 1

The sum is $$\Im\sum_{r=0}^\infty\left(\frac13\exp i\theta\right)^r=\Im\frac{1}{1-\frac13\exp i\theta}=\Im\frac{1-\frac13\exp -i\theta}{\frac{10}{9}-\frac23\cos\theta}=\frac{3\sin\theta}{10-6\cos\theta}.$$For $\sin\theta=\frac13$, $\cos\theta=\pm\frac{2\sqrt{2}}{3}$, so the sum simplifies to $\frac{1}{10\mp 4\sqrt{2}}=\frac{5\pm 2\sqrt{2}}{34}$.

Answer 2

Use geometric series. We have $$\begin{split}\sum_{r=0}^\infty\frac{\sin(r\theta)}{3^r} &=\sum_{r=0}^\infty\mathrm{Im}\left(\frac{e^{ir\theta}}{3^r}\right) \\&=\mathrm{Im}\left(\sum_{r=0}^\infty\left(\frac{e^{i\theta}}{3}\right)^r\right)\\&=\mathrm{Im}\left(\frac{1}{1-e^{i\theta}/3}\right)\\&=\mathrm{Im}\left(\frac{1+e^{-i\theta}/3}{1+(e^{i\theta}-e^{-i\theta})/3-1/9}\right) \\&=\mathrm{Im}\left(\frac{1+\cos(\theta)/3-i\sin(\theta)/3}{8/9+2\sin(\theta)/3}\right)\\&=\frac{-1/9}{8/9+2/9}\\&=-\frac{1}{10}\end{split}$$ if I didn't make a silly mistake somewhere.