# Sum of this series

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

Hints in the right direction would be appreciated.

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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 have 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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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 : $$\sum_{n=0}^{+\infty} \frac{\sin^3(3^n)}{3^n} = \frac{3}{4} \sin(1)$$

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