Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 79



Work Step by Step

Given: $3+\frac{9}{2!}+\frac{27}{3!}+....=\Sigma_{n=1}^{\infty}\dfrac{3^{n}}{(n!)}$ The above series resembles: $e^{x}-1=\Sigma_{n=1}^{\infty}\dfrac{x^{n}}{(n!)}$ Thus, $\Sigma_{n=1}^{\infty}\dfrac{3^{n}}{(n!)}=e^{3}-1$
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