Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.2 - Infinite Series - Exercises 10.2 - Page 579: 15

Answer

$\dfrac{5}{3}$

Work Step by Step

Here, $a=1, r=\dfrac{2}{5}$; Since, $r=|\dfrac{2}{5}| \lt 1$; Thus, the series converges. Formula to find the sum of a geometric series is $S=\dfrac{a}{1-r}$ Thus, $S=\dfrac{1}{1-\dfrac{2}{5}}=\dfrac{5}{3}$
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