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 497: 15



Work Step by Step

Here, $a=1, r=\dfrac{2}{5}$; The common ratio is $|\dfrac{2}{5}| \lt 1$; so the series converges. The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}=\dfrac{1}{1-\dfrac{2}{5}}=\dfrac{5}{3}$
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