## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 11 - Section 11.6 - Absolute Convergence and the Ratio and Root Tests - 11.6 Exercises - Page 743: 41

#### Answer

By the root test, the series absolutely converges.

#### Work Step by Step

$\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}\frac{b^{n}_{n}cos(n\pi)}{n}$ Replace $cosn \pi$ with $(-1)^{n}$ $\sum_{n=1}^{\infty}\frac{b^{n}_{n}cos(n\pi)}{n}=\sum_{n=1}^{\infty}\frac{b^{n}_{n}(-1)^{n}}{n}$ $|a_{n}|=|\frac{b^{n}_{n}(-1)^{n}}{n}|=\frac{b^{n}_{n}}{n}$ $\lim\limits_{n \to \infty} \sqrt[n] {|a_{n}|}=\lim\limits_{n \to \infty} \sqrt[n] {\frac{b^{n}_{n}}{n}}$ $=\lim\limits_{n \to \infty} {\frac{b_{n}}{n^{1/n}}}$ $= {\dfrac{\lim\limits_{n \to \infty}b_{n}}{\lim\limits_{n \to \infty}n^{1/n}}}$ $=\frac{1/2}{1}$ $=\frac{1}{2}\lt 1$ By the root test, the series absolutely converges.

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