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: 20



Work Step by Step

We have: $0.234234234= \dfrac{234}{1000}+\dfrac{234}{(1000)^2}+\dfrac{234}{(1000)^3}+....=\dfrac{234}{1000}(1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..)$ The sum of a geometric series $1+ \dfrac{1}{1000}+ \dfrac{1}{(1000)^2}+..$ can be found as: $S=\dfrac{a}{1-r}$ Here, $a=1, r=\dfrac{1}{1000}=0.001$; Thus, $S=\dfrac{234}{1000}(\dfrac{1}{1-0.001})=\dfrac{26}{111}$
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