Calculus: Early Transcendentals (2nd Edition)

by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard

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: 44

Answer

$=\dfrac {5}{9}$

Work Step by Step

$0.\overline {5}=0.555...=\dfrac {5}{10}+\dfrac {5}{10^{2}}+\dfrac {5}{10^{3}}+\dfrac {}{..}=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {5}{10}}{1-\dfrac {1}{10}}=\dfrac {\dfrac {5}{10}}{\dfrac {9}{10}}=\dfrac {5}{9}$

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