Answer
The series is divergent.
Work Step by Step
$\Sigma^{\infty}_{n=1} \frac{1+3^{n}}{2^{n}}$
$s_{n} =\Sigma^{\infty}_{n=1} a_{n}$
$=\Sigma^{\infty}_{n=1} \frac{1+3^{n}}{2^{n}}$
$=\Sigma^{\infty}_{n=1} \frac{1}{2^{n}} + \Sigma^{\infty}_{n=1}\frac{3^{n}}{2^{n}}$
$=\Sigma^{\infty}_{n=1} \frac{1}{2^{n}} + \Sigma^{\infty}_{n=1} (\frac{3}{2})^{n}$
For $\Sigma^{\infty}_{n=1} \frac{1}{2^{n}}$ the series can be expressed as $\frac{1}{2}, \frac{1}{4}, \frac{1}{6}...$
The first term $a=\frac{1}{2}$ and the common ratio $r= \frac{1}{2} \lt 1$
Since $|r| \lt 1$, the geometric series $=\Sigma^{\infty}_{n=1} \frac{1}{2^{n}}$ is convergent and the sum is
$\Sigma^{\infty}_{n=1} \frac{1}{2^{n}} = \frac{a}{1-r}$
$=\frac{\frac{1}{2}}{1-\frac{1}{2}}$
$=\frac{\frac{1}{2}}{\frac{1}{2}}$
$=1$
For $\Sigma^{\infty}_{n=1} (\frac{3}{2})^{n}$, the series can be expressed as $\frac{3}{2}, \frac{9}{4}, \frac{27}{8}...$
The first term is $a=\frac{3}{2}$ amd the common ratio is $r=\frac{3}{2} \gt 1$
Since $|r| \gt 1$, the geometric series $=\Sigma^{\infty}_{n=1} (\frac{3}{2})^{n}$ is divergent.
If $\Sigma a_{n}$ is convergent and $\Sigma b_{n}$ is divergent, then the sum is a divergent series.
