Study the sequence $x_n=\sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}}$. I am given the sequence:
$$
x_n=\sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}}
$$
Where $n \ge 2$. I have to describe the sequence $(x_n)_{n \ge 2}$, where I am given the following options (I have to choose one):
A. convergent
B. bounded and divergent
C. unbounded and divergent
D. has negative terms
E. has infinite limit
What I tried was to use something like:
$$y_n = \ln x_n$$
$$y_n = \dfrac{\ln(2^{n \sin 1} + 2^{n \sin 2} + 2^{n \sin 3} + ... + 2^{n \sin n})}{n}$$
But I got stuck here and I can't find the limit. I should probably use a different approach, but I can't think of any. I can't continue what I've started either.
How should I approach this?
By the way, is choice B just something to trick me, or is it an actual possibility? Can a sequence be bounded and divergent? That doesn't seem right...
 A: Let
$$
a_n=\max\{\sin k: k=1,\ldots,n\}.
$$
Then $\{a_n\}$ is increasing and $a_n\le 1$, for all $n\in\mathbb N$, and hence convergent to some positive $a\le 1$. (In fact, it converges to 1.)
Then
$$
2^{na_n}\le 2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}\le n\cdot 2^{na_n}
$$
and hence
$$
2^{a_n}= \sqrt[n]{2^{na_n}}\le \sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}}\le 2^{a_n}\sqrt[n]{n}
$$
But $\,\,2^{a_n}\to 2^a$ and $\,\,\sqrt[n]{n}2^{a_n}\to 2^a$, and hence
$$
\sqrt[n]{2^{na_n}}\le \sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}} \to 2^a.
$$
Thus A is the correct answer.
Note. In fact $a=1$, and thus 
$$
\sqrt[n]{2^{n\sin 1}+2^{n\sin 2}+\cdots+2^{n\sin n}}\to 2.
$$
This is due to Weyl's Equidistribution Criterion.
A: not a solution 
Here is $n=2$ to $n=1000$.

Visual impression: converges to $2$, but not monotonically.
A: As  @Jam mentioned in comments and GEdgar in his answer, it is not monotonic at all as shown in the table for the very first values of $n$ (put them on a scatter plot)
$$\left(
\begin{array}{cc}
 n & x_n \\
 1 & 1.79188 \\
 2 & 2.59580 \\
 3 & 2.39393 \\
 4 & 2.22146 \\
 5 & 2.12745 \\
 6 & 2.07215 \\
 7 & 2.07967 \\
 8 & 2.19713 \\
 9 & 2.16297 \\
 10 & 2.13347
\end{array}
\right)$$
To extend GEdgar plot, consider $n=10^k$ and the values are
$$\left(
\begin{array}{cc}
 k & x_{10^k} \\
 1 & 2.133472777 \\
 2 & 2.030634143 \\
 3 & 2.005381446 \\
 4 & 2.000775098 \\
 5 & 2.000100408 \\
 6 & 2.000012379
\end{array}
\right)$$
From these numbers, it seems that, more or less,
$$\log(x_n-2)\sim 0.567936 -1.96092\, k \qquad (R^2=0.998712)$$
