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: 8



Work Step by Step

Here, $a=\dfrac{1}{16}, r=\dfrac{1}{4}$ The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}=\dfrac{\dfrac{1}{16}}{1-\dfrac{1}{4}}=\dfrac{\dfrac{1}{16}}{\dfrac{3}{4}}=\dfrac{1}{12}$
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