Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.4 - The Comparison Tests - 11.4 Exercises - Page 731: 27



Work Step by Step

Use Limit Comparison Test with $a_{n}=(1+\frac{1}{n})^{2}e^{-n}$ and $b_{n}=e^{-n}$ $\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{(1+\frac{1}{n})^{2}e^{-n}}{e^{-n}}$ $=\lim\limits_{n \to \infty}{(1+\frac{1}{n})^{2}}$ $=(1+0)^{2}\ne 1 \ne 0 \ne \infty $ $\Sigma_{n=1}^{\infty}e^{-n}$ is convergent by the Limit Comparison Test and $\Sigma e^{-n}$ converges because it is a geometric series with $|r|\lt 1$.
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