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: 57

Answer

Series converges, and the sum is $\frac{-5x}{1+5x}$

Work Step by Step

$\Sigma^{\infty}_{n=1} (-5)^{n}x^{n} = \Sigma^{\infty}_{n=1}(-5x)^{n}$ $a=-5x$ and $r=-5x$ $|r| \lt 1$ $|-5x| \lt 1$ $|5x| \lt 1$ $-\frac{1}{5} \lt x \lt \frac{1}{5}$ Therefore, the series converges, and the sum is $\Sigma^{\infty}_{n=1} (-5x)^{n} = \frac{a}{1-r}$ $=\frac{-5x}{1+5x}$
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