## Calculus: Early Transcendentals 8th Edition

Given: $a_n=\frac{cos^{2}n}{2^{n}}$ Note that we can write $\frac{0}{2^{n}} \leq \frac{cos^{2}n}{2^{n}}\leq \frac{1}{2^{n}}$ Also, $0 \leq \frac{cos^{2}n}{2^{n}}\leq \frac{1}{2^{n}}$ Now, $\lim\limits_{n \to \infty}\frac{1}{2^{n}}=\frac{1}{\infty}=0$ Therefore, by Sandwich Theorem $\lim\limits_{n \to \infty}\frac{cos^{2}n}{2^{n}}=0$ Hence, the given sequence converges to $0$.