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: 13

Answer

Divergent

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: $\Sigma_{n=1}^{\infty}(-1)^{n-1}e^{2/n}$ $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}(-1)^{n-1}e^{2/n}=(-1)^{n-1}= DNE$ As $n$ increases, $2/n$ approaches to $0$. Because of the alternating sign, the limit will oscillate between $-1$ and $1$. The limit is not zero. Hence, the given series diverges by the divergence test.
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