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 497: 12

Answer

$\dfrac{17}{2}$

Work Step by Step

We have two geometric series: $(5-1)+(\dfrac{5}{2} -\dfrac{1}{3})+(\dfrac{5}{4} -\dfrac{1}{9})+(\dfrac{5}{8} -\dfrac{1}{27})+...$ Here, $a=5, r=\dfrac{1}{2}$ and $a=1, r=\dfrac{1}{3}$ The sum of a geometric series can be found as: $S=s_1 -s_2=\dfrac{a}{1-r}$ or, $=\dfrac{5}{1-\dfrac{1}{2}}+\dfrac{1}{1-\dfrac{1}{3}}=10+ - \dfrac{3}{2}=\dfrac{17}{2}$
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