The entire question is essentially given in the title: For $ n $ a positive integer, does the sequence $ x_n = \sqrt[n]{\left\lvert \sin(2^n) \right\rvert} $ have a limit as $ n \to \infty $?
Background: One of the questions in a recent batch of homework I marked was to investigate the convergence or divergence of the infinite series $$ \sum_{n=1}^{\infty} \frac{\sin(2^n)}{2^n} \, . $$ Obviously, this series absolutely converges by comparison since $$ \left\lvert \frac{\sin(2^n)}{2^n} \right\rvert \leq \frac{1}{2^n} \, , $$ and $ \sum_{n=1}^{\infty} \frac{1}{2^n} $ is a convergent geometric series, with common ratio $ r = \frac{1}{2} $. This is the the solution provided in the solution sheet, and essentially the way most people solved the question.
However, one script attempted to use the root test to determine whether or not the series converged. So, let $ a_n = \frac{\sin(2^n)}{2^n} $, then one needs to determine the limit as $ n \to \infty $ of $$ \sqrt[n]{\left\lvert a_n \right\rvert} = \sqrt[n]{\left\lvert \frac{\sin(2^n)}{2^n} \right\rvert} = \frac{1}{2} \sqrt[n]{ \left\lvert \sin(2^n) \right\rvert } \, , $$ which brings me to the question in the title. Does the sequence $ x_n = \sqrt[n]{ \left\rvert \sin(2^n) \right\rvert } $ have a limit, or not?
I can't convince myself either way, whether this sequence has a limit or not. Since there is a much better way of determining the convergence of the given infinite series, it wasn't worth spending too much time figuring out whether or not this solution worked. But I am still curious about it, so can anyone resolve this question for me, one way or the other?
Thoughts: If it were $ f(x) = \sqrt[x]{ \left\lvert \sin(2^x) \right\rvert } $, with $ x > 0 $, then I can easily see $ \lim_{x \to \infty} f(x) $ does not exist. Just look at the subsequences $ f(\log_2(n\pi)) = 0 $, and $ f(\log_2(n\pi + \frac{1}{2}\pi)) = 1 $, which have limits $ 0 $ and $ 1 $ respectively, as $ n \to \infty $.
Restricting to $ x_n = f(n) $, $ n $ a positive integer, seems a much more difficult question. My instinctive reaction is to say $ \sin(2^n) $ is such a 'chaotic'/badly behaved sequence, that $ x_n $ can't possibly have a limit. But this is nowhere near a proof, and I can't justify it to myself any better than this. On the other hand, some numerical computations with Maple suggest the sequence might be approaching the limit $ 1 $.
I guess the question of whether this sequence has a limit is tied strongly to how close $ 2^n $ gets to a rational multiple of $ \pi $, putting us in the realm of Diophantine approximation (about which I know little). More specifically, what bounds in terms of $ n $ can be put on the difference $ \left\lvert 2^n - b \pi \right\rvert $. In this vein, I have been told that the function $ \sin(2^n) $ is dense in $ [0,1] $, and that this is not obvious. But by itself I don't think this is enough to conclude one way or another. Heuristically: depending on how soon $ \sin(2^n) $ gets close to 0, the $ n $-th root may be enough to force this close to 1 anyway.