Answer
32
Work Step by Step
Testing the ratios of neighboring terms,
$\displaystyle \frac{6}{8}=\frac{3}{4}$
$\displaystyle \frac{\frac{9}{2}}{6}=\frac{9}{12}=\frac{3}{4}$
$\displaystyle \frac{\frac{27}{8}}{\frac{9}{2}}=\frac{27}{8}\div\frac{9}{2}=\frac{27}{8}\cdot\frac{2}{9}=\frac{3}{4}$
we see that this is a geometric series,
$a_{1}=8, |r|=|\displaystyle \frac{3}{4}| < 1,$ so by Th.9.6,
$\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}$
$\displaystyle \sum_{n=0}^{\infty}8(\frac{3}{4})^{n}=\frac{8}{1-(\frac{3}{4})}=\frac{8}{\frac{1}{4}}=8\cdot 4=32$
