Answer
$\frac{9}{11}$
Work Step by Step
$0.\overline {81}=0.818181\ldots =\dfrac {81}{100}+\dfrac {81}{100^2}+\dfrac {81}{100^3}\ldots =\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {81}{100}}{1-\dfrac {1}{100}}=\dfrac {81}{99}=\dfrac {9}{11}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 9 - Infinite Series - 9.2 Exercises - Page 601: 37
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