Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 744: 46


Converges to $0$

Work Step by Step

Given: $a_n=({2})^{-n}cos \pi n$ It can be re-written as $a_n=\frac{cos n \pi}{{2}^{n}}$ $cos n \pi$ is a bounded function. $-1 \leq cos n\pi\leq 1$ Place upper and lower bounds into inequality. $\lim\limits_{n \to \infty}\frac{-1}{2^{n}} \leq \lim\limits_{n \to \infty}\frac{cos n \pi}{{2}^{n}}\leq \lim\limits_{n \to \infty}\frac{1}{2^{n}}$ $0 \leq \lim\limits_{n \to \infty}\frac{cos n \pi}{{2}^{n}}\leq 0$ Therefore, by Squeeze Theorem the given sequence converges to $0$.
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