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: 45

Answer

$=\dfrac {1}{11}$

Work Step by Step

$0.\overline {09}=0.090909...=\dfrac {9}{100}+\dfrac {9}{100^{2}}+\dfrac {9}{100^{3}}+...=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {9}{100}}{1-\dfrac {1}{100}}=\dfrac {\dfrac {9}{100}}{\dfrac {99}{100}}=\dfrac {9}{99}=\dfrac {1}{11}$
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