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: 12



Work Step by Step

Alternating series test: Suppose that we have a 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 conditions are satisfied, the series is convergent. 1. $\lim\limits_{n \to \infty}b_{n}=0$ 2. $b_{n}$ is a decreasing sequence. Given: $\Sigma_{n=1}^{\infty}(-1)^{n+1}ne^{-n}$ In the given problem, $b_{n}=ne^{-n}=\frac{ n}{e^{n}}$ which satisfies both conditions of Alternating Series Test as follows: 1. $b_{n}=\frac{ n}{e^{n}}$ is decreasing because the denominator is increasing. 2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{ n}{e^{n}}$ Since the limit is in the form of $\frac{\infty}{\infty}$, we can use L-hospital's rule. $=\lim\limits_{n \to \infty}\frac{ 1}{e^{n}}$ $=\frac{1}{\infty}$ $=0$ Hence, the given series is convergent by the Alternating Series Test.
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