Evalute $\sum_{n=0}^{\infty}\dfrac{\sin(2^n x)}{2^n}$ Can this be expressed in a closed form?
$$f(x)=\sum_{n=0}^{\infty}\dfrac{\sin(2^n x)}{2^n}$$
I believe it converges as $-1\le\sin x\le 1$, but I am not sure whether it is even possible to evaluate it in a closed form. I tried to evaluate $f'(x)$, which I failed. I tried to transform the series into a telescoping one, which I also failed.

You can compare this sequence to a certain famous one to establish
convergence:
$$
\sum_{n=0}^{\infty}\frac{\sin(2^{n}x)}{2^{n}}\leq\sum_{n=0}^{\infty}\left|\frac{\sin(2^{n}x)}{2^{n}}\right|\leq\sum_{n=0}^{\infty}\frac{1}{2^{n}}=2
$$
However, I doubt it has a nice solution in terms of elementary functions.
If you are interested in computing the actual value of the series
at different points in $x$, you can compute the $N$-th partial sums for a large enough $N$. The results of such a computation are

Note that only $0\leq x\leq 2\pi$ is shown since the function is periodic.
