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

Hints in the right direction would be appreciated.

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up vote 19 down vote accepted

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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