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: 23

Answer

$\dfrac{1}{15}$

Work Step by Step

Here, $0.06666=\dfrac{6}{100}(1+ \dfrac{1}{10}+ \dfrac{1}{(10)^2}+..)$ Formula to find the sum of a geometric series$1+ \dfrac{1}{10}+ \dfrac{1}{(10)^2}+..$ is $S=\dfrac{a}{1-r}$ Since we have , $a=1, r=\dfrac{1}{10}=0.1$; Then , we get $S=\dfrac{6}{100}(\dfrac{1}{1-0.1})=\dfrac{1}{15}$
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