Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.3 Infinite Series - 8.3 Exercises - Page 623: 43


$=\dfrac {1}{9}$

Work Step by Step

$0.\overline {1}=0.1111...=\dfrac {1}{10}+\dfrac {1}{10^{2}}+\dfrac {1}{10^{3}}+\ldots =\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {1}{10}}{1-\dfrac {1}{10}}=\dfrac {\dfrac {1}{10}}{\dfrac {9}{10}}=\dfrac {1}{9}$
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