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: 39

Answer

$=\dfrac {9}{460}\approx 0.019565$

Work Step by Step

$\sum ^{\infty }_{k=2}\left( -0.15\right) ^{k}=\dfrac {a_{1}}{1-r};a_{1}=\left( -0.15\right) ^{2}=0.0225;r=\left( -0.15\right) \Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {0.0225}{1-\left( -0.15\right) }=\dfrac {225}{11500}=\dfrac {9}{460}\approx 0.019565$
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