## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.4 Absolute and Conditional Convergence - Exercises - Page 563: 10

#### Answer

The series $\Sigma_{n=1}^{\infty}\frac{\cos n}{2^n}$ converges absolutley.

#### Work Step by Step

Since $\cos n\leq 1$ then we have $\frac{\cos n}{2^n}\leq \frac{1}{2^n}$. Now, the series $\Sigma_{n=1}^{\infty}\frac{1}{2^n}$ is a geometric series with $r=1/2\lt1$, which converges. Hence, the positive series $\Sigma_{n=1}^{\infty}|\frac{\cos n}{2^n}|$ converges. Thus, the series $\Sigma_{n=1}^{\infty}\frac{\cos n}{2^n}$ converges absolutley.

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