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 - Page 716: 19


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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.