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: 26

Answer

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

Work Step by Step

Consider the series as: $3.\overline{142857}=3+\Sigma_{n=0}^\infty \dfrac{142857}{10^6}\dfrac{1}{(10^6)^n}$ : Formula to calculate the sum of a geometric series is $S=\dfrac{a}{1-r}$ Then, we have $S=3+\dfrac{\dfrac{142857}{10^6}\dfrac{1}{(10^6)^n}}{1-10^6}$ or, $S=\dfrac{116,402}{37,037}$ or, $\dfrac{22}{7}$
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