Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.10 Exercises - Page 790: 69

Answer

$e^{3}-1$

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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