Calculus 10th Edition

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: 36

Answer

$\frac{4}{11}$

Work Step by Step

$0.\overline {36}=0.363636\ldots =\dfrac {36}{100}+\dfrac {36}{100^2}+\dfrac {36}{100^3}\ldots =\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {36}{100}}{1-\dfrac {1}{100}}=\dfrac {36}{99}=\dfrac {4}{11}$
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