Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises: 19

Answer

Convergant $=\dfrac {25}{3}$

Work Step by Step

$10-2+0.4-0.08=\sum ^{\infty }_{k=0}10\times \left( -\dfrac {1}{5}\right) ^{k}\Rightarrow \lim _{n\rightarrow \infty }10\times \left( -\dfrac {1}{5}\right) ^{k}=0$ $\sum ^{\infty }_{k=0}\left( 10\times \left( -\dfrac {1}{5}\right) ^{k}\right) -\dfrac {a_{1}}{1-r};a_{1}=10;r=\left( -\dfrac {1}{5}\right) \Rightarrow S_{\infty }=\dfrac {a_{1}}{1-r}=\dfrac {10}{1-\left( -\dfrac {1}{5}\right) }=\dfrac {25}{3}$
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