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: 26


$\dfrac{116,402}{37,037}$ or, $\dfrac{22}{7}$

Work Step by Step

We have: $3.\overline{142857}=3+\Sigma_{n=0}^\infty \dfrac{142857}{10^6}\dfrac{1}{(10^6)^n}$ The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ Thus, $S=3+\dfrac{\dfrac{142857}{10^6}\dfrac{1}{(10^6)^n}}{1-10^6}=\dfrac{116,402}{37,037}$ or, $\dfrac{22}{7}$
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