Is there a way to determine $\sum_{n=0}^{\infty}\sin(2^{n}\theta )$? Is there way to determine/simplify this infinite sum below?
$$\sum_{n=0}^{\infty}\sin(2^{n}\theta )$$
 A: For any $\theta \in \mathbb{R}$, consider the binary representation of the fractional part of $\displaystyle\;\frac{\theta}{\pi}\;$
$$\left\{ \frac{\theta}{\pi} \right\} = (0.b_1b_2b_3\ldots)_2 =
\sum_{n=1}^\infty \frac{b_n}{2^n}$$
If the bit change between the $(n+1)^{th}$ and $(n+2)^{th}$ slot, i.e. $b_{n+1}b_{n+2} = 01\text{ or }10$, then
$$\left\{ \frac{2^n\theta}{\pi} \right\} = (0.b_{n+1}b_{n+2}\ldots)_2 \in \left[\frac14,\frac34\right] \implies |\sin(2^n\theta)| \ge \frac{1}{\sqrt{2}}$$
This means if the bits $b_n$ change infinitely often, there is no chance for $\sin(2^n\theta)$ tend to $0$.
As a result, $\sum\limits_{n=0}^\infty \sin(2^n\theta)$ diverges.
If the bits $b_n$ only change finitely many times, $b_n$ will settle to either $0$ or $1$ for sufficiently large $n$. For such $n$, $2^n\theta$ becomes an integer multiple of $\pi$. The series at hand become a finite sum and hence converges.
Conclusion: $\sum\limits_{n=0}^\infty\sin(2^n\theta)$ converges when any only when $\displaystyle\;\theta = \frac{m\pi}{2^n}\;$ for some $m \in \mathbb{Z}, n \in \mathbb{N}$.
