Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Practice Exercises - Page 636: 23



Work Step by Step

The sum of the geometric series is given by $s_n=\dfrac{a}{(1-r)}$ Here, $a$ is initial term and $r$ is common ratio. A geometric series is to be convergent when $|r|\lt 1$ and divergent when $|r|\gt 1$. From the given problem, we have $r=e^{-1} \lt 1$ Thus, $s_n=\dfrac{a}{(1-r)}=\dfrac{1}{(1-e^{-1})}$ Hence, $s_n=\dfrac{e}{(e-1)}$
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