Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 716: 63

Answer

The series converges when $x \lt 0$ and the sum is $\frac{1}{1-e^{x}}$.

Work Step by Step

$\Sigma^{\infty}_{n=0}e^{nx}$ $a=1$ and $r=e^{x}$ $|r| \lt 1$ $|e^{x}| \lt 1$ $e^{x} \lt 1$ $x \lt ln_{e}1$ $x \lt 0$ Hence the series converges if $x \lt 0$ $\Sigma^{\infty}_{n=0}e^{nx} = \frac{a}{1-r}$ $=\frac{1}{1-e^{x}}$
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