I was trying to come up with an interesting example (not like $(-1)^n$ or cyclic like $1,2,3, 1,2,3, 1,2,3, \ldots$) of a bounded not converging sequence with an explicit convergent subsequence.
I though of something like $x_n := \sin(n)$. Since $\sin(k\pi) = 0$ for every $k \in \mathbb{Z}$ I thought that the closer the argument of the sine was to a multiple of $\pi$, the closer sine of that argument should be to zero (continuity...). Therefore, the sequence $x_n := \sin(a_n \cdot 10^{n - 1})$, where $a_n$ are the first $n$ digits of $\pi$ should be converging to $0$, right? If yes, how can I show it?
$\lim\limits_{n\rightarrow \infty} |a_n10^{n-1}-10^{n-1}\pi| \neq 0 $ because for any large $N$ you can find $n>N$ such that this difference will be bigger $0.1$.
You are looking for $a_n-\pi$ in this case you will have monotonic convergence