Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.2 Exercises - Page 737: 74


$\ln \frac{9}{10}$

Work Step by Step

$\Sigma^{\infty}_{n=1} e^{nc} =\Sigma^{\infty}_{n=1}(e^{c})^{n}$ $a_{1} = e^{0} =1$ and the common ratio is $r=e^{c}$ $\frac{1}{1-e^{c}}=10$ $1-e^{c} = \frac{1}{10}$ $-e^{c} = -\frac{9}{10}$ $e^{c} = \frac{9}{10}$ $c=\ln \frac{9}{10}$
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.