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: 17

Answer

$\dfrac{1}{7}$

Work Step by Step

The Nth partial sum of a geometric series can be found as: $s_n=\dfrac{a(1-r^n)}{1-r}$ The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ The common ratio for a geometric series is $a=\dfrac{1}{8}, r=\dfrac{1}{8}$; Thus, $S=\dfrac{\dfrac{1}{8}}{1-\dfrac{1}{8}}=\dfrac{1}{7}$; converges
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