Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 9 - Infinite Series - Review Exercises - Page 677: 35

Answer

$\dfrac{1}{11}$

Work Step by Step

The sum of a geometric series can be found as: $S_n=\dfrac{a_1}{1-r}$ where, $a_1$ denotes the first term and the $r$ is common ratio. We are given that $\overline{0.09}=0.09+0.0009+....=\dfrac{9}{10^2}+\dfrac{9}{10^4}+...$ This can be further written as: $\dfrac{9}{10^2}+\dfrac{9}{10^4}+...=\Sigma_{n=0}^{\infty} \dfrac{9}{10^2} (\dfrac{1}{10^2})^n=\Sigma_{n=0}^{\infty} \dfrac{9}{100} (\dfrac{1}{10^2})^n$ So, we have $a_1=\dfrac{9}{100}$ and $r=\dfrac{1}{100}$ Thus, the total sum of two geometric series is: $S_n=\dfrac{9/100}{1-\dfrac{1}{100}}=\dfrac{1}{11}$
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