Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - Review - True-False Quiz: 10



Work Step by Step

Since, $e^{x}=\Sigma_{n=0}^{\infty}\frac{x^{n}}{n!}$ Put the value of $x$ into the series as $\frac{1}{e}=e^{-1}$, so we can plug in $x=-1$ Therefore, $e^{-1}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{n!}$ OR $\frac{1}{e}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{n!}$ Hence, the statement is TRUE.
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