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?

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 $$\frac {\sin (3)} {4} $$.
