Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - Chapter Review Exercises - Page 591: 29


$$ \frac{4}{3}$$

Work Step by Step

Given $$\frac{4}{9}+\frac{8}{27}+\frac{16}{81}+\frac{32}{243}+\cdots$$ This is a geometric series with $|r|= \frac{8}{27}\frac{9}{4}=\frac{2}{3}<1$, so the series converges and has the sum \begin{align*} S&=\frac{a}{1-r}\\ &=\frac{4/9}{1-\frac{2}{3}}\\ &= \frac{4}{3} \end{align*}
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