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: 20

Answer

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

Work Step by Step

$\sum ^{\infty }_{k=0}\left( \dfrac {3}{5}\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=\left( \dfrac {3}{5}\right) ^{0}=1;r=\dfrac {3}{5}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {1}{1-\dfrac {3}{5}}=\dfrac {5}{2}=2.5$
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