Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.2 - Infinite Series - Exercises 10.2 - Page 579: 13

Answer

$\dfrac{17}{6}$

Work Step by Step

Consider two geometric series: $(1+1)+(\dfrac{1}{2} -\dfrac{1}{5})+(\dfrac{1}{4} +\dfrac{1}{25})+(\dfrac{1}{8} -\dfrac{1}{125})+...$ Here, $a=5, r=\dfrac{1}{2}$ and $a=1, r=\dfrac{-1}{5}$ Formula to find the sum of a geometric series is $S=\dfrac{a}{1-r}$ or, $S=s_1 +s_2=\dfrac{1}{1-\dfrac{1}{2}}+\dfrac{1}{1+\dfrac{1}{5}}$ or, $S=2+(\dfrac{5}{6})=\dfrac{17}{6}$
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