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: 49


$2+\sqrt 2$

Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ Here, $a=1, r=\dfrac{1}{\sqrt 2} \lt 1$ Thus, $S=\dfrac{a}{1-r}=\dfrac{1}{1-\dfrac{1}{\sqrt 2}}=\dfrac{\sqrt 2}{\sqrt 2-1}=2+\sqrt 2$
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