Evaluating $\sum_{n=0}^{\infty}{\sin^{3}\left(3^{n}\right) \over 3^{n}}$ 
How do I find the sum of this series?
  $$\sum_{n=0}^{\infty}{\sin^{3}\left(3^{n}\right) \over 3^{n}}$$

Hints in the right direction would be appreciated.
 A: From the identity $ \displaystyle \left(\sin(x)\right)^3 = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^3$  you get $ \displaystyle \left(\sin(x)\right)^3 = \frac{3}{4}\sin(x)-\frac{1}{4}\sin(3x) $
In particular : $ \displaystyle \frac{\sin^3(3^n)}{3^n} = \frac{3}{4}\left(\frac{\sin(3^n)}{3^n}-\frac{\sin(3^{n+1})}{3^{n+1}}\right)$ 
So we have a telescopic partial sum : $$ \sum_{n=0}^N \frac{\sin^3(3^n)}{3^n} = \frac{3}{4} \sin(1)-\frac{3}{4} \frac{\sin(3^{N+1})}{3^{N+1}}  $$
So since $ \displaystyle \lim_{x \rightarrow +\infty} \frac{\sin(x)}{x} = 0 $  by taking the limit as $ N  \rightarrow +\infty $ gives : $$ \sum_{n=0}^{+\infty} \frac{\sin^3(3^n)}{3^n} = \frac{3}{4} \sin(1) $$
A: Hint: Use this 
$\sin^3x=\frac{3}{4}\sin x - \frac{1}{4}\sin3x. $
Solution:
Let $a_n=\frac{\sin^33^n}{3^n}$ and $b_n=\frac{\sin3^n}{3^n}$. So from the identity mentioned above, one simply has that
$
a_n = \frac{1}{4}(3b_n -3b_{n+1}).
$
Therefore,
$$
\sum_{n\ge 0} \frac{\sin^33^n}{3^n} = \sum_{n\ge 0} a_n = \sum_{n\ge 0} \frac{1}{4}(3b_n -3b_{n+1}) = \frac{3}{4}b_0 =  \frac{3\sin 1}{4}.
$$
