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: 19



Work Step by Step

We have: $0.232323= \dfrac{23}{100}+\dfrac{23}{(100)^2}+\dfrac{23}{(100)^3}+....=\dfrac{23}{100}(1+ \dfrac{1}{100}+ \dfrac{1}{(100)^2}+..)$ The sum of a geometric series $1+ \dfrac{1}{100}+ \dfrac{1}{(100)^2}+..$ can be found as: $S=\dfrac{a}{1-r}$ We have: $a=1, r=\dfrac{1}{100}=0.01$; Thus, $S=\dfrac{23}{100}(\dfrac{1}{1-0.01})=\dfrac{23}{99}$
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