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: 2

Answer

Convergent

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: $\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}-....$ General Term $a_{n}=(-1)^{n+1}\frac{2}{2n+1}$ Thus, $\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}-....=\Sigma_{n=1}^{\infty}(-1)^{n+1}\frac{2}{2n+1}$ In the given problem, $b_{n}=\frac{2}{2n+1}$ which satisfies both conditions of Alternating Series Test as follows: 1. Let $f(x)=\frac{2}{2x+1}$ and $f'(x)=\frac{-2.2}{(2x+1)^{2}} \lt 0$ Thus, $f(x)$ is deceasing , so $b_{n}=\frac{2}{2n+1}$ is decreasing because the denominator is increasing. 2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{2}{2n+1}=\frac{2}{\infty}=0$ Hence, the given series is convergent by Alternating Series Test.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.