## Calculus (3rd Edition)

$36$
The series $\sum_{n=-1}^{\infty} \frac{2^{n+3}}{3^{n}}$ can be rewritten as follows $$\sum_{n=-1}^{\infty} \frac{2^{n+3}}{3^{n}}=\sum_{n=-1}^{\infty} 8\left(\frac{2}{3}\right)^{n}= 8\left(\frac{2}{3}\right)^{-1}+ \sum_{n=0}^{\infty} 8\left(\frac{2}{3}\right)^{n}.$$ Now, we have a geometric series $\sum_{n=0}^{\infty} 8\left(\frac{2}{3}\right)^{n}$ which has the sum $$\frac{c}{1-r}=\frac{8}{1-(2/3)}=24.$$ Hence, the sum of the series $\sum_{n=-1}^{\infty} \frac{2^{n+3}}{3^{n}}$ is $$12+24=36.$$