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: 4



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: $\frac{1}{ln3}-\frac{1}{ln4}+\frac{1}{ln 5}-\frac{1}{ln 6}+\frac{1}{ln 7}-....$ General Term $a_{n}=(-1)^{n+1}\frac{1}{ln(n+2)}$ Thus, $\frac{1}{ln3}-\frac{1}{ln4}+\frac{1}{ln 5}-\frac{1}{ln 6}+\frac{1}{ln 7}-.....=\Sigma_{n=1}^{\infty}(-1)^{n+1}\frac{1}{ln(n+2)}$ In the given problem, $b_{n}=\frac{1}{ln(n+2)}$ which satisfies both conditions of Alternating Series Test as follows: 1. $b_{n}=\frac{1}{ln(n+2)}$ is decreasing because the denominator is increasing. 2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{1}{ln(n+2)}=\frac{1}{\infty}=0$ Hence, the given series is convergent by Alternating Series Test.
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