Answer
$\Sigma$$\frac{n+3^{n}}{n+2^{n}}$ is divergent
Work Step by Step
Set
$a_{n}$ = $\frac{n+3^{n}}{n+2^{n}}$
$b_{n}$ = $\frac{3^{n}}{2^{n}}$ = $\frac{3}{2}$^{n}
$\lim\limits_{n \to \infty}$ $\frac{a_{n}}{b_{n}}$ = $\lim\limits_{n \to \infty}$ $\frac{\frac{n}{3^{n}}+1}{\frac{n}{2^{n}}+1}$
Then solve the two small limits respectively:
$\lim\limits_{n \to \infty}$ $\frac{n}{3^{n}}$ = $\lim\limits_{n \to \infty}$ $\frac{1}{n \times 3^{n-1}}$ (L'Hopital rule) = 0
$\lim\limits_{n \to \infty}$ $\frac{n}{2^{n}}$ = $\lim\limits_{n \to \infty}$ $\frac{1}{n \times 2^{n-1}}$ (L'Hopital rule) = 0
Therefore
$\lim\limits_{n \to \infty}$ $\frac{a_{n}}{b_{n}}$ = $\lim\limits_{n \to \infty}$ $\frac{\frac{n}{3^{n}}+1}{\frac{n}{2^{n}}+1}$ = $\frac{0+1}{0+1}$ = 1 > 0
since
$\lim\limits_{n \to \infty}$ $\frac{a_{n}}{b_{n}}$ > 0
$\Sigma$$b_{n}$ is divergent ($\frac{3}{2}$ > 1 , $b_{n}$ is geometric series)
then
$\Sigma$$\frac{n+3^{n}}{n+2^{n}}$ is divergent by limit comparison test.
