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

Question

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 $\dfrac{1}{27\cos(2\theta)}$.

Answer 1

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.

Answer 2

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

By $\sin3\theta =3\sin \theta-4\sin^3\theta$

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

Then

$$\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}\cdot4\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}\left[3\sin\left(\frac{\pi}{3^{n+1}}\right)-\sin\left(\frac{3\pi}{3^{n+1}}\right)\right]$$

$$=\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}\left[3\sin\left(\frac{\pi}{3^{n+1}}\right)-\sin\left(\frac{3\pi}{3^{n+1}}\right)\right]$$

$$=\frac{1}{4}\sum_{n=0}^\infty \left[3^n\sin\left(\frac{\pi}{3^{n+1}}\right)-3^{n-1}\sin\left(\frac{3\pi}{3^{n}}\right)\right]$$

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

$$\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}\cdot4\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\sum_{n=0}^m\left[ f(n)-f(n-1)\right]$$

$$=\sum_{n=0}^m\left[ f(n)-f(n-1)\right]=f(m)-f(-1)$$

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

$$=\frac{3^m}{4}\sin\left(\frac{\pi}{3^{m+1}}\right)-0$$

Now see $$=\frac{3^m}{4}\sin\left(\frac{\pi}{3^{m+1}}\right)= \frac{\pi}{12} \cdot \left(\frac{\sin \frac{\pi}{3^{m+1}}}{\frac{\pi}{3^{m+1}}}\right)$$

Now take the limit $m \to \infty$.