Is there a close form solution for this series? I want to find close form solution for this series
$$\sum_{n=1}^{\infty}\cos\left(\frac{x}{3^n}\right)\sin^2\left(\frac{x}{3^n}\right)$$
Thanks for your help.
 A: Hint
Set
$$f(x)=\sum\limits_{n=1}^{+\infty }{{{3}^{n-1}}{{\sin }^{3}}\left( \frac{x}{{{3}^{n}}} \right)}$$
we have
$$f'(x)=\sum\limits_{n=1}^{+\infty }{\cos \left( \frac{x}{{{3}^{n}}} \right){{\sin }^{2}}\left( \frac{x}{{{3}^{n}}} \right)}$$
Now use this identity and find $f(x)$
$${{\sin }^{3}}\theta =\frac{3}{4}\sin \theta -\frac{1}{4}\sin 3\theta $$ 
A: Hint:
Linearise each term:
\begin{align*}\cos\left(\frac{x}{3^n}\right)\sin^2\left(\frac{x}{3^n}\right)&=\cos\left(\frac{x}{3^n}\right)\times\frac12\biggl(1-\cos\Bigl(\frac{2x}{3^n}\Bigr)\biggr)=\frac12\biggl(\cos\Bigl(\frac{x}{3^n}\Bigr)-\cos\Bigl(\frac{x}{3^n}\Bigr)\cos\Bigl(\frac{2x}{3^n}\Bigr)\biggr)\\
\\&=\frac12\biggl(\cos\Bigl(\frac{x}{3^n}\Bigr)-\frac12\biggl(\cos\Bigl(\frac{3x}{3^n}\Bigr)+\cos\Bigl(\frac{x}{3^n}\Bigr)\biggr)\biggr)\\
&=\frac14\biggl(\cos\Bigl(\frac{x}{3^n}\Bigr)-\cos\Bigl(\frac{x}{3^{n-1}}\Bigr)\biggr).
\end{align*}
You now have a telescoping series.
A: Playing around with the series numerically (plotting it etc) it seems it's equal to $\frac12 \sin^2(\frac{x}{2})$. I don't have a proof right now (and I'm too lazy to work one out), but it seems to work. Please feel free to check it yourself ;-)
With this in mind, you might be able to come up with something.
Cheers!
