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: 22



Work Step by Step

We have: $0.dddd= \dfrac{d}{10}+\dfrac{d}{(10)^2}+\dfrac{d}{(10)^3}+....=\dfrac{d}{10}(1+ \dfrac{1}{10}+ \dfrac{1}{(10)^2}+..)$ The sum of a geometric series $1+ \dfrac{1}{10}+ \dfrac{1}{(10)^2}+..$ can be found as: $S=\dfrac{a}{1-r}$ Here, $a=1, r=\dfrac{1}{10}=0.01$; Thus, $S=\dfrac{d}{10}(\dfrac{1}{1-0.01})=\dfrac{d}{9}$
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