Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 547: 29



Work Step by Step

Given $$ \sum_{n=1}^{n=\infty} \left(e\right)^{-n}= \sum_{n=1}^{n=\infty} \left(\frac{1}{e}\right)^{-n}$$ Since the series is a geometric series with $|r|=\left|\frac{1}{e}\right|<1$, then the series converges and has the sum \begin{align*} S&= \frac{a_1}{1-r}\\ &=\frac{\frac{1}{e}}{1-\frac{1}{e}}\\ &=\frac{1}{1-e} \end{align*}
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