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: 27


$=\dfrac {1}{7}$

Work Step by Step

$\sum ^{\infty }_{n=1}\left( 2^{-3k}\right) =\sum ^{\infty }_{k=1}8^{-k}=\dfrac {a_{1}}{1-r};a_{1}=8^{-1};r=\dfrac {1}{8}\Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {\dfrac {1}{8}}{1-\dfrac {1}{8}}=\dfrac {1}{7}$
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