University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 59



Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ Since, we have $\sum_{n =0}^{ \infty} (\dfrac{2}{3})^n$ and $\sum_{n =0}^{ \infty} (\dfrac{1}{3})^n$ We have a convergent geometric series with first term, $a=1$ and common ratio $r =\dfrac{2}{3}, \dfrac{1}{3}$ We find the difference between the sums of the two geometric series thus: Thus, $S=\dfrac{1}{1-\dfrac{2}{3}}-\dfrac{1}{1-\dfrac{1}{3}}=\dfrac{3}{3-1}=\dfrac{3}{2}$
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