## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 744: 43

#### Answer

converges to $0$

#### Work Step by Step

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 $0$.

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