<s>[INST] 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.
 [/INST] 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,
$$
\sum_{n\ge 0} \frac{\sin^33^n}{3^n} = \sum_{n\ge 0} a_n = \sum_{n\ge 0} \frac{1}{4}(3b_n -3b_{n+1}) = \frac{3}{4}b_0 =  \frac{3\sin 1}{4}.
$$
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