TITLE: A community project: prove (or disprove) that $\sum_{n\geq 1}\frac{\sin(2^n)}{n}$ is convergent
QUESTION [100 upvotes]: As the title says, I would like to launch a community project for
  proving that the series $$\sum_{n\geq 1}\frac{\sin(2^n)}{n}$$ is
  convergent.

An extensive list of considerations follows. The first fact is that the inequality
$$ \sum_{n=1}^{N}\sin(2^n)\ll N^{1-\varepsilon}\qquad\text{or}\qquad\sum_{n=1}^{N}e^{2^n i}\ll \frac{N}{\log(N)^{1+\varepsilon}} \tag{1}$$
for some $\varepsilon>0$ is enough to prove the claim by Abel summation. In the same spirit, it is quite common to employ Weyl's inequality / Van der Corput's trick to prove the convergence of $\sum_{n\geq 1}\frac{\sin(n^p)}{n}$. In our case, however, we do not have an additive base of $\mathbb{N}$ made by perfect powers associated with some exponent, hence an additive base of finite order, so the estimation of the exponential sums appearing in the right side of $(1)$ is more difficult. Assuming that for an infinite number of primes $p$ the element $2$ is a generator of $\mathbb{Z}/(p\mathbb{Z})^*$ (Legendre's conjecture), we may probably regard $[1,N]$ as a subset of $\mathbb{Z}/(p\mathbb{Z})$ (for a huge $p$) and prove there is enough cancellation to grant $(1)$. However, Legendre's conjecture seems quite out-of-reach at the moment.
Integral extimations techniques, that turned out to be pretty effective in other contexts, are almost ineffective here, since $\sin(2^x)$ oscillates too fast, so that there is no reason for expecting that
$$ \sum_{n=M+1}^{M+N}\sin(2^n)\approx \int_{M}^{N+M}\sin(2^x)\,dx, $$
so, even if $\lim_{N\to +\infty}\int_{1}^{N}\sin(2^x)\,dx$ is convergent by Dirichlet's test, there is little use of that.
However, the series $g(\xi)=\sum_{n\geq 1}\frac{\sin(2^n\xi)}{n}$ is convergent for almost every $\xi$, since $g\in L^2\left(-\pi,\pi\right)$.
Additionally, the statement there is enough cancellation to ensure $(1)$ appears to be equivalent to (or, at least, a consequence of) the statement the digits $0$ and $1$ in the binary representation of $\pi$ are equidistributed. It is hard to believe that is not the case, and the Bailey-Borwein-Plouffe formula
$$\pi=\sum_{k\geq 0}\frac{1}{16^k}\left(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6}\right)\tag{2}$$
may provide a way to prove it.

REPLY [2 votes]: This may be of help...
Let $cis(x)=\cos(x)+i\sin(x)$
Since $2^n$ are never complex, $\Re(cis(2^n))=\cos(2^n)$ and $\Im(cis(2^n)=i\cdot sin(2^n)$ then we can say 
$f=\sum_{n\geq 1}\frac{\sin(2^n)}{n}$ diverges $\iff \Im(g)$ diverges
Where $\displaystyle g=\sum_{n\geq 1}\frac{cis(2^n)}{n}$
$\displaystyle g=\sum_{n=1}^{\infty} \frac{e^{2^ni}}{n}$
So the question is equivalent to whether the following converges:
$\Im\left(\displaystyle\frac{e^{2i}}{1}+\displaystyle\frac{e^{4i}}{2}+\displaystyle\frac{e^{8i}}{3}+\displaystyle\frac{e^{16i}}{4}+\cdots\right)$