$\sum_{n=0}^{\infty} \frac{\sin(nx)}{3^n}$

Question

I'm currently doing AOPS vol.2 and this problem has stumped me for weeks.

Find $$\sum_{n=0}^{\infty} \frac{\sin(nx)}{3^n}$$ Knowing:$$\sin (x)=\frac{1}{3}$$ $$0\leq x\leq \frac{\pi}{2}$$

My Attempt: Using$$\sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}$$ and plugging that back into the series but i cant seem to reduce it anymore. The book has not talked about $\ln$ or $\log$ trig, so that will not be useful. But it has discussed Euler's formulas such as $e^{ix}=\cos(x)+i\sin(x)$ and the aforementioned $\sin(x)$.

Please answer using no to a little amount of calculus as I've not studied too far into it (Taylor series etc). Your solution will be much appreciated.

Answer 1

Note that$$\sum_{n=0}^\infty\frac{\sin(nx)}{3^n}=\operatorname{Im}\left(\sum_{n=0}^\infty\frac{e^{inx}}{3^n}\right)=\operatorname{Im}\left(\sum_{n=0}^\infty\left(\frac{e^{ix}}3\right)^n\right)=\operatorname{Im}\left(\frac1{1-\frac{e^{ix}}3}\right).$$Can you take it from here?

Answer 2

$$\sum_{n\ge0}\frac{\sin (nx)}{3^n}=\frac{1}{2i}\sum_{n\ge0}\left[\left(\frac13 e^{ix}\right)^n-\left(\frac13 e^{-ix}\right)^n\right]=\frac{1}{2i}\left[\frac{1}{1-\frac13 e^{ix}}-\frac{1}{1-\frac13 e^{-ix}}\right]\\=\frac{1}{2i}\frac{e^{ix}-e^{-ix}}{\frac{10}{3}-2\cos x}=\frac{3\sin x}{2(5-3\cos x)}.$$Using $\sin x=\frac13,\,\cos x=\sqrt{1-\sin^2x}=\frac{2\sqrt{2}}{3}$, the above simplifies to$$\frac{1}{2(5-2\sqrt{2})}=\frac{5+2\sqrt{2}}{34}.$$