A convergent series: $\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$ I would like to find the value of:

$$\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$$

I could only see that the ratio of two consecutive terms is $\dfrac{1}{27\cos(2\theta)}$.

Hint. One may observe that
$$
\sin^3 (a)=\frac34\sin (a)-\frac14 \sin(3a)
$$ giving

$$
3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\frac{3^{n}}4\sin\left(\frac{\pi}{3^{n+1}}\right)-\frac{3^{n-1}}4\sin\left(\frac{\pi}{3^{n}}\right)
$$ 

then one gets a telescoping sum.
