University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 75


Converges to $\dfrac{1}{2+x}$ for all $|x+1| \lt 1$ or, $-2 \lt x \lt 0$

Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ We have a convergent geometric series with first term, $a=1$ and common ratio $r =-(x+1)^n$ $S=\dfrac{1}{1-(-(x+1)^n)}=\dfrac{1}{2+x}$ Hence, the given series converges to $\dfrac{1}{2+x}$ for all $|x+1| \lt 1$ or, $-2 \lt x \lt 0$
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