Let $f(x)$ denote the sum of the infinite trigonometric series, $f(x) =\sum^{\infty}_{n=1} \sin\frac{2x}{3^n}\sin\frac{x}{3^n}$ Let $f(x)$ denote the sum of the infinite trigonometric series, 
$$f(x) =\sum^{\infty}_{n=1} \sin\frac{2x}{3^n} \sin\frac{x}{3^n}$$
I am not getting any clue how to solve it only thing I can put the value of $n = 1$, $2$, $3$, $\infty$. Please help thanks.

Hints:
$$\sin\alpha\sin\beta=\frac12\left(\cos(\alpha-\beta)-\cos(\alpha+\beta)\right)\implies$$
$$\sin\frac{2x}{3^n}\sin\frac x{3^n}=\frac12\left(\cos\frac x{3^n}-\cos\frac x{3^{n-1}}\right)\implies$$
$$\sum_{n=1}^\infty\sin\frac{2x}{3^n}\sin\frac x{3^n}=\frac12\left(\cos\frac x3-\cos x+\cos\frac x9-\cos\frac x3+\cos\frac x{27}-\cos\frac x9+\ldots\right)$$
Can you see the telescopic series...?
