Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.5 - Alternating Series - 11.5 Exercises - Page 736: 7



Work Step by Step

Alternating series test: Suppose that we have series $\Sigma a_n$, such that $a_{n}=(-1)^{n}b_n$ or $a_{n}=(-1)^{n+1}b_n$, where $b_n\geq 0$ for all $n$. Then if the following two condition are satisfied the series is convergent. 1. $\lim\limits_{n \to \infty}b_{n}=0$ 2. $b_{n}$ is a decreasing sequence. Given: $\Sigma_{n=1}^{\infty}(-1)^{n}\frac{3n-1}{2n+1}$ In the given problem, $b_{n}=\frac{3n-1}{2n+1}$ 1. Test the first condition of the AST as: $b_{n+1}\leq b_n$ $\frac{3(n+1)-1}{2(n+1)+1}\leq \frac{3n-1}{2n+1}$ $\frac{3n+2}{2n+3}\leq \frac{3n-1}{2n+1}$ $(3n+2)(2n+1)\leq (3n-1)(2n+3)$ $2\leq -3$ which is not true, so AST first condition fails. 2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{3n-1}{2n+1}=\lim\limits_{n \to \infty}\frac{3-1/n}{2+1/n}=\frac{3}{2}\ne 0$ which means that the series diverges by the Test of Divergence.
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