Answer
The series is convergent and the sum of series is $\frac{3e}{3-e}$.
Work Step by Step
$\Sigma^{\infty}_{n=0} \frac{e^{n}}{3^{n-1}} = \Sigma^{\infty}_{n=0} \frac{e \times e^{n-1}}{3^{n-1}}$
$=\Sigma^{\infty}_{n=0} e(\frac{e}{3})^{n-1}$
$a=e$ and $r=\frac{e}{3}$
Since $2 \lt e \lt 3$
Therefore $\frac{e}{3} \lt 1$
$r \lt 1$
The series is convergent
$\Sigma^{\infty}_{n=0} \frac{e^{n}}{3^{n-1}} = \frac{a}{1-r}$
$= \frac{e}{1-\frac{e}{3}}$
$=\frac{3e}{3-e}$
