Only Two Terms Remains, Which Terms?

Consider the following definition :

$f_k(x) = \sum_{n=1}^k \sin \frac {2x}{3^n} \sin \frac {x}{3^n}$

The identity $2\sin x \sin y = \cos (x-y) - \cos (x+y)$ prompts us to telescope the above sum to get the following result : $2f_k(x) = \cos \frac{x}{3^k} - \cos x$

Taking $k \to \infty$ and noting that $2\sin^2 x = 1 - \cos 2x$ , we get $f(x) = \sin^2 \left( \frac{x}{2} \right)$

Let $\text{S} = \sum_{n=1}^{\infty} \sin \left( \dfrac{2x}{3^n}\right) \sin \left( \dfrac{x}{3^n} \right)$

$= \dfrac{1}{2} \sum_{n=1}^{\infty} 2\sin \left( \dfrac{2x}{3^n}\right) \sin \left( \dfrac{x}{3^n} \right)$

$= \dfrac{1}{2} \sum_{n=1}^{\infty} \left(\cos \left( \dfrac{x}{3^n} \right) - \cos \left( \dfrac{x}{3^{n-1}} \right)\right)$

By noting that the above sum telescopes, we have,

$\text{S} = \lim_{n \to \infty} \left[ \dfrac{1}{2} \left( \cos\left( \dfrac{x}{3^n} \right) - \cos x \right) \right]$

$= \dfrac{1}{2} (1-\cos x)$

Making the answer $\boxed{4}$ .