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: 25



Work Step by Step

Here, $1.24123123123=\dfrac{124}{100}+\dfrac{123}{100000}(1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..)$ Formula to find the sum of a geometric series$(1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..)$ is $S=\dfrac{a}{1-r}$ Since, we have $a=1, r=\dfrac{1}{1000}=0.001$; Then , we get $S=\dfrac{124}{100}+\dfrac{123}{100000}(\dfrac{1}{1-0.001})=\dfrac{41,333}{33,300}$
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