Sequence limit! Think carefully..

At first, one may think that it always has limit one since $\sin(2\pi\gamma n)$ is never zero for irrational $\gamma$ and $\frac{1}{n^{n}}$ approach zero very fast. But actually we can find some $\gamma$ such that there is a subsequence of $\{\sin(2\pi\gamma n)\}$ approach zero so fast that the corresponding subsequence of $\{|\sin(2\pi\gamma n)|^{\frac{1}{n^{n}}}\}$ also goes to zero. Following is a way to construct: Let $\gamma=0.10..010....010...$ where the $\gamma$ is consists of digits $0,1$ and the number of consecutive $0$ after the $n$ th $1$ is given by $10^{(k_{n}+1)10^{k_{n}}}$ , where $k_{n}$ is the position where we encounter $n$ th $1$ in the decimal expression of $\gamma$ . For example, after the first $1$ appears, there are $10^{20}$ zeroes after it. $k_{1}=1, k_{n+1}=k_{n}+10^{(k_{n}+1)10^{k_{n}}}+1$ which increase to infinity extremely fast. Now because $\sin(2\pi\gamma n)=\sin(2\pi\{\gamma n\})$ where $\{x\}=x-[x]$ and $[x]$ is the integer part of $x$ , $|\sin(2\pi\{\gamma 10^{k_{n}}\})|\leq 2\pi\{\gamma 10^{k_{n}}\}\leq 2\pi*2*10^{-(k_{n+1}-k_{n})}\leq 2\pi*10^{-10^{(k_{n}+1)10^{k_{n}}}}$ . Then the $10^{k_{n}}$ th term of the sequence is less then $(2\pi*10^{-10^{(k_{n}+1)10^{k_{n}}}})^{\frac{1}{10^{k_{n}10^{k_{n}}}}}=(2\pi)^{\frac{1}{10^{k_{n}10^{k_{n}}}}}10^{-10^{10^{k_{n}}}}$ which goes to zero!

And also for any value between -1, 1 , there is a subsequence of $\sin(2\pi\gamma n)$ has the limit of that value (intuitively true but may not that easy to prove ). If the value is not zero, then the corresponding subsequence of $\{|\sin(2\pi\gamma n)|^{\frac{1}{n^{n}}}\}$ goes to 1. So the sequence doesn't converge.