Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.3 Infinite Series - 8.3 Exercises - Page 623: 26


$=\dfrac {5}{2}=2.5$

Work Step by Step

$\sum ^{\infty }_{m=2}\dfrac {5}{2^{m}}=\dfrac {a_{1}}{1-r};a_{1}=\dfrac {5}{2^{2}}=\dfrac {5}{4};r=\dfrac {1}{2}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {5}{4}}{1-\dfrac {1}{2}}=\dfrac {5}{2}=2.5$
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