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: 29


$=\dfrac {1}{500}=0.002$

Work Step by Step

$\sum ^{\infty }_{k=4}\dfrac {1}{5^{k}}=\dfrac {a_{1}}{1-r};a_{1}=\dfrac {1}{5^{4}};r=\dfrac {1}{5}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {1}{5^{4}}}{1-\dfrac {1}{5}}=\dfrac {1}{500}=0.002$
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