## Thomas' Calculus 13th Edition

$\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{\sin^2 n}{2^n}$ Since, we know that $-1 \leq \sin n \leq 1$ , thus apply Sandwich Theorem, which states that $\lim\limits_{n \to 0} \dfrac{1}{n}=0$ and $-1 \leq \sin^2 n \leq 1$ So, $\lim\limits_{n \to \infty}\dfrac{-1}{2^n}=0$ This implies that, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{\sin^2 n}{2^n}=0$ Thus, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent