Sum of series $\sum \limits_{k=1}^{\infty}\frac{\sin^3 3^k}{3^k}$ Calculate the following sum: $$\sum \limits_{k=1}^{\infty}\dfrac{\sin^3 3^k}{3^k}$$
Unfortunately I have no idea how to handle with this problem.
Could anyone show it solution?
 A: Using 
$$\sin(3a)=3\sin a-4\sin^3a \to \color{red}{\sin^3(a)=\frac14\Big(3\sin a-\sin(3a)\Big)} $$
so
\begin{eqnarray}
\sum_{k=1}^{\infty}\frac{\sin^3(3^k)}{3^k}
&=&
\frac14\sum_{k=1}^{\infty}\frac{3\sin(3^k)-\sin(3.3^k)}{3^k}\\
&=&
\frac14\sum_{k=1}^{\infty}\frac{\sin(3^k)}{3^{k-1}}-\frac{\sin(3^{k+1})}{3^{k}}\\
&=&
\frac14\sum_{k=1}^{\infty}f(k)-f(k+1)\\
&=&\frac14\Big(\frac{\sin3}{3^{1-1}}-\lim_{n \to \infty}\frac{\sin(3^{n+1})}{3^n}\Big)\\
&=&\frac{\sin(3)}{4}
\end{eqnarray}
A: We can use the following identity (leaving for you to prove)
$$
\sin^3 (3^x) = \frac {1}{4} ( 3 \sin(3^x) - \sin (3^{x+1}) )
$$
That is,
$$
\frac {\sin^3( 3) }{3} =  \frac {1}{4} ( \sin (3) - \frac {1}{3} \sin (3^2) )
$$
$$
\frac {\sin^3( 3^2) }{3^2} =  \frac {1}{4} (\frac {1}{3} \sin (3^2) - \frac {1}{3^2} \sin (3^3) )
$$
$$
\frac {\sin^3( 3^3) }{3^3} =  \frac {1}{4} (\frac {1}{3^3} \sin (3^3) - \frac {1}{3^4} \sin (3^4) )
$$
As you can clearly see, all the middle terms cancel out in the sum and we are only left with the first term. So the sum is $$\frac {\sin (3)} {4} $$.
A: An overkill. Let $\mathfrak{M}\left(*,s\right)
 $ the Mellin transform. Using the identity $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\lambda_{k}g\left(\mu_{k}x\right),\, s\right)=\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}\mathfrak{M}\left(g\left(x\right),s\right)
 $$ we have $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\frac{\sin^{3}\left(3^{k}x\right)}{3^{k}},\, s\right)=\underset{k\geq1}{\sum}\left(\frac{1}{3^{s+1}}\right)^{k}\mathfrak{M}\left(\sin^{3}\left(x\right),s\right)
 $$ and since $$\mathfrak{M}\left(\sin^{3}\left(x\right),s\right)=\frac{\Gamma\left(s\right)\sin\left(\frac{\pi}{2}s\right)}{4}\left(3-\frac{1}{3^{s}}\right)
 $$ we have for $\textrm{Re}\left(s\right)>-1
 $ $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\frac{\sin^{3}\left(3^{k}x\right)}{3^{k}},\, s\right)=\frac{\Gamma\left(s\right)\sin\left(\frac{\pi}{2}s\right)}{4}\left(3-\frac{1}{3^{s}}\right)\frac{1}{3^{s+1}-1}=\frac{\Gamma\left(s\right)\sin\left(\frac{\pi}{2}s\right)3^{-s}}{4}
 $$ and so inverting we get $$\underset{k\geq1}{\sum}\frac{\sin^{3}\left(3^{k}x\right)}{3^{k}}=\frac{1}{2\pi i}\int_{1/2-i\infty}^{1/2+i\infty}\frac{\Gamma\left(s\right)\sin\left(\frac{\pi}{2}s\right)3^{-s}}{4}x^{-s}ds
 $$ now taking $x=1$ and shifting the complex line to the left we have, from the residue theorem, that we have to evaluate the residues of $$\textrm{Res}_{s=-2n-1} \frac{\Gamma\left(s\right)\sin\left(\frac{\pi}{2}s\right)3^{-s}}{4}=\frac{\left(-1\right)^{k}}{4\left(2k+1\right)!3^{-2n-1}}
 $$ hence $$\sum_{k\geq1}\frac{\sin^{3}\left(3^{k}\right)}{3^{k}}=\frac{1}{4}\sum_{k\geq 0}\frac{\left(-1\right)^{k}}{\left(2k+1\right)!}3^{2k+1}=\color{red}{\frac{\sin\left(3\right)}{4}}.$$
