Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 809: 86

Answer

$e^{3/5}$

Work Step by Step

Given: $\Sigma_{n=0}^{\infty}\dfrac{3^{n}}{5^{n}(n!)}$ which can be written as $\Sigma_{n=0}^{\infty}\dfrac{(3/5)^{n}}{n!}$ As we know: $e^{x}=\Sigma_{n=0}^{\infty}\dfrac{x^{n}}{(n!)}$ Then, $e^{3/5}=\Sigma_{n=0}^{\infty}\dfrac{3^{n}}{5^{n}(n!)}$
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