Answer
$\displaystyle \frac{27}{4}$
Work Step by Step
Testing the ratios of neighboring terms,
$\displaystyle \frac{-3}{9}=-\frac{1}{3}$
$\displaystyle \frac{1}{-3}=-\frac{1}{3}$
$\displaystyle \frac{-\frac{1}{3}}{1}=-\frac{1}{3}$
we see that this is a geometric series,
$a=9, |r|=|-\displaystyle \frac{1}{3}| < 1,$ so by Th.9.6,
$\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$
$\displaystyle \sum_{n=0}^{\infty}9(-\frac{1}{3})^{n}=\frac{9}{1-(-\frac{1}{3})}=\frac{9}{\frac{4}{3}}=\frac{9\cdot 3}{4}= \frac{27}{4}$
