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: 57



Work Step by Step

The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ Thus, we have $\sum_{n =1}^{ \infty} \dfrac{2}{10^n}$ We have a convergent geometric series with first term, $a=\dfrac{2}{10}$ and common ratio $r =\dfrac{1}{10}$ Thus, $S=\dfrac{\dfrac{2}{10}}{1-\dfrac{1}{10}}=\dfrac{2}{9}$
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