## University Calculus: Early Transcendentals (3rd Edition)

$\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent
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