Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 716: 51

Answer

$=\dfrac {8}{9}$

Work Step by Step

$0.\overline {8}=0.8888\ldots =\dfrac {8}{10}+\dfrac {8}{10^{2}}+\dfrac {8}{10^{3}}...=\dfrac {a_{1}}{1-r};a_{1}=\dfrac {8}{10};r=\dfrac {1}{10}\Rightarrow 0.\overline {8}=\dfrac {\dfrac {8}{10}}{1-\dfrac {1}{10}}=\dfrac {8}{9}$
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