A community project: prove (or disprove) that $\sum_{n\geq 1}\frac{\sin(2^n)}{n}$ is convergent 
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.
 A: 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:
$\Im\left(\displaystyle\frac{e^{2i}}{1}+\displaystyle\frac{e^{4i}}{2}+\displaystyle\frac{e^{8i}}{3}+\displaystyle\frac{e^{16i}}{4}+\cdots\right)$
A: References:
[1] N.Bary: A Treatise on Trigonometric Series, Volume 1 & 2.
[2] A.Zygmund: Trigonometric Series, Volume 1 & 2.
[3] P.Erdos, S.Taylor: On The Set of Points of Convergence of a Lacunary Trigonometric Series and the Equidistribution Properties of Related Sequences.
[4] A.Zygmund: On Lacunary Trigonometric Series.
[5] JP.Kahane: Geza Freud and Lacunary Fourier Series.
[6] JP.Kahane: Lacunary Taylor and Fourier Series.

Discussion:
Following to the remark regarding the convergent of the series { $\sum_{n \geq 1} \sin(\xi \space 2^{n})/n$ } for almost every $\xi$ , Let { $\xi = 2^{m} \space\colon\space m \in \Bbb N$ } and define the function $f(m)$:
$$ \boxed{ f(m) = \sum_{n=1}^{\infty} \frac{\sin(2^{m} \space 2^{n})}{n} } \\[8mm] $$
$$
\begin{align}
f(m-1)-f(m) & = \sum_{n=1}^{\infty} \left[ \frac{\sin(2^{m-1} \space 2^{n})}{n} - \frac{\sin(2^{m} \space 2^{n})}{n} \right] \\[4mm]
& = \small \left[ \frac{\sin(2^{m} 2^{0})}{1} - \frac{\sin(2^{m} 2^{1})}{1} \right] + \left[ \frac{\sin(2^{m} 2^{1})}{2} - \frac{\sin(2^{m} 2^{2})}{2} \right] + \left[ \frac{\sin(2^{m} 2^{2})}{3} - \text{...} \right] + \text{...} \\[4mm]
& = \small \frac{\sin(2^{m} 2^{0})}{1} - \left[ \frac{\sin(2^{m} 2^{1})}{1} - \frac{\sin(2^{m} 2^{1})}{2} \right] - \left[ \frac{\sin(2^{m} 2^{2})}{2} - \frac{\sin(2^{m} 2^{2})}{3} \right] - \text{...} \\[4mm]
& = \sin(2^{m}) - \sum_{n=1}^{\infty} \frac{\sin(2^{m} \space 2^{n})}{n(n+1)} \\
\end{align}
$$
Applying summation by parts:
$$ f(0)-f(N) = \small \left[ f(0)-f(1) \right] + \left[ f(1)-f(2) \right] + \text{...} + \left[ f(N-1)-f(N) \right] = \normalsize \sum_{m=1}^{N} \left[ f(m-1)-f(m) \right] \\[6mm]
= \sum_{m=1}^{N} \sin(2^m) - \sum_{m=1}^{N} \sum_{n=1}^{\infty} \frac{\sin(2^m \space 2^n)}{n(n+1)} = \sum_{m=1}^{N} \sin(2^m) - \sum_{n=1}^{\infty} \frac{\sum_{m=1}^{N} \sin(2^m 2^n)}{n(n+1)} $$
Which implies:
$$ \boxed{ \sum_{n=1}^{N} \sin(2^n) \quad\text{bounded}\quad \iff f(0)-f(N) = \sum_{n=1}^{\infty} \frac{\sin(2^{n}) - \sin(2^{n+N})}{n} \quad\text{convergent}\quad } $$
And the question is equivalent to show $|f(0)-f(N)|$ is convergent. Although it is not so clear how to argue the boundary, at least it is a result of subtracting two series with equally divergent speed and same term limit. As well as, if $f(m)$ converge for a value of $m$, then $f(m)$ converge for all values of $m$ and vies versa (implying the initial remark for the special case $\xi = 2^{m}$).
If $\sum \sin(2^{n})/n$ convergent then  $\sum \sin(2^{n+N})/n$ convergent too, and everything is okay. On the other hand, assuming $\sum \sin(2^{n})/n$ divergent then $\sum \sin(2^{n+N})/n$ divergent too. And because of the equally speed and same limit, the subtracting "$\small \underline{\text{is potentially}}$" convergent, resulting in $\sum \sin(2^{n})/n$ convergent (false assumption). The cosine case is the same (with interest).
$$ \sum_{n=1}^{N} \cos(2^n) = \mathcal{O}(1) \iff \sum_{n=1}^{\infty} \frac{\cos(2^{n}) - \cos(2^{n+N})}{n} = 2\sum_{n=1}^{\infty} \frac{\cos^{2}(2^{n-1}) - \cos^{2}(2^{n-1+N})}{n} $$
