Does $ \sqrt[n]{\left\lvert \sin(2^n) \right\rvert} $ have a limit?

Despite starting out as a question of convergence of series, this seems to be an issue of Diophantine approximation.

Instead of looking at $2^n (\text{mod } \pi)$ look at the binary expansion $\frac{1}{\pi}2^n (\text{mod } 1)$. Then I found this result in a paper and this blog:

Lemma: There exists orbit closures $\{2^nx\}_{n\geq0}$ of any Hausdorff dimension between 0 and 1.

Theorem:(Furstenberg) The orbit closure $\{2^m 3^n x\}_{m,n\geq0}$ is $\mathbb{R}/\mathbb{Z}$ or finite, according to whether $x$ is irrational or rational.

More information on the dynamics of the map $x \mapsto 2x (\text{mod }1)$ is covered in this Master's thesis of Johan Nilsson, Fractal Structures in Dyadic Diophantine Approximation. An example of such a fractals set might be $\{ x: \{2^n x\} < \frac{3}{4} \text{ for all } n\in \mathbb{N} \}$.