Answer
The series is convergent and the sum is $\frac{5}{2}$
Work Step by Step
$\Sigma^{\infty}_{n=1}ar^{n-1} = \frac{a}{1-r}$ and $|r| \lt 1$
If $|r| \geq 1$, then the geometric series is divergent
$\Sigma^{\infty}_{n=1} \frac{1+2^{n}}{3^{n}}$
$\Sigma^{\infty}_{n=1} (\frac{1}{3^{n}} + \frac{2^{n}}{3^{n}}) = \Sigma^{\infty}_{n=1} \frac{1}{3^{n}} + \Sigma^{\infty}_{n=1} \frac{2^{n}}{3^{n}}$
$=\Sigma^{\infty}_{n=1} (\frac{1}{3})^{n} + \Sigma^{\infty}_{n=1}(\frac{2}{3})^{n}$
First
$\Sigma^{\infty}_{n=1} (\frac{1}{3})^{n} = \Sigma^{\infty}_{n=1} \frac{1}{3} (\frac{1}{3})^{n-1}$
$a= \frac{1}{3}$ and the common ratio is $r=\frac{1}{3}$
$|r|=|\frac{1}{3}|$
$=\frac{1}{3}$
$=0.3333$
Since $|r| \lt 1$, then $\Sigma^{\infty}_{n=1} (\frac{1}{3})^{n}$ is convergent and the sum is
$\Sigma^{\infty}_{n=1} (\frac{1}{3})^{n} = \frac{a}{1-r}$
$=\frac{\frac{1}{3}}{1-\frac{1}{3}}$
$=\frac{\frac{1}{3}}{\frac{2}{3}}$
$=\frac{1}{2}$
Second
$\Sigma^{\infty}_{n=1}(\frac{2}{3})^{n} = \Sigma^{\infty}_{n=1}\frac{2}{3}(\frac{2}{3})^{n-1}$
$a=\frac{2}{3}$ and the common ratio is $r=\frac{2}{3}$
$|r| =|\frac{2}{3}|$
$=\frac{2}{3}$
$=0.66666$
Since $|r| \lt 1$ the geometric series $\Sigma^{\infty}_{n=1}(\frac{2}{3})^{n}$ is convergent with the sum
$\Sigma^{\infty}_{n=1}(\frac{2}{3})^{n} = \frac{a}{1-r}$
$=\frac{\frac{2}{3}}{1-\frac{2}{3}}$
$=\frac{\frac{2}{3}}{\frac{1}{3}}$
$=2$
If $\Sigma a_{n}$ and $\Sigma b_{n}$ are convergent series then the sum is a convergent series.
The sum of the convergent series is $\frac{1}{2} + 2 = \frac{5}{2}$
