Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 757: 76

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

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