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: 74

Answer

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

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 $r=e^{c}$ $\frac{1}{1-e^{c}}=10$ $1-e^{c} = \frac{1}{10}$ $-e^{c} = -\frac{9}{10}$ $e^{c} = \frac{9}{10}$ $c=\ln \frac{9}{10}$
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