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: 71


Converges to $\dfrac{6}{3-x}$ for $ -1 \lt \dfrac{x-1}{2} \lt 1$ or, $ -1 \lt x \lt 3$

Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ The given series shows a convergent geometric series with first term, $a=3$ and common ratio $r =\dfrac{x-1}{2}$ $S=\dfrac{3}{1-(\dfrac{x-1}{2})}=\dfrac{6}{3-x}$ Hence, the given series converges to $\dfrac{6}{3-x}$ for $ -1 \lt \dfrac{x-1}{2} \lt 1$ or, $ -1 \lt x \lt 3$
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