Answer
$\lim\limits_{n \to \infty} a_n=0 $ and {$a_n$} is convergent
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\dfrac{\sin n}{n}$
Since, $-1 \leq \sin n \leq 1$ and $\lim\limits_{n \to \infty} \dfrac{1}{n}=0$, we apply the Sandwich Theorem.
Thus, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}
\dfrac{\sin n}{n}=0$
Hence, $\lim\limits_{n \to \infty} a_n=0 $ and {$a_n$} is convergent
