Answer
Divergent
Work Step by Step
$\Sigma^{\infty}_{n=1}(\frac{3}{5^{n}} + \frac{2}{n})$
$=\Sigma^{\infty}_{n=1} \frac{3}{5^{n}} + \Sigma^{\infty}_{n=1} \frac{2}{n}$
$= 3\Sigma^{\infty}_{n=1} \frac{1}{5^{n}} + 2\Sigma^{\infty}_{n=1} \frac{1}{n}$
First
$\Sigma^{\infty}_{n=1} \frac{1}{5^{n}} = \frac{1}{5} + \frac{1}{5^{2}}+\frac{1}{5^{3}}+\frac{1}{5^{4}}+...$
The common ratio $r= \frac{1}{5} \lt 1$
Therefore $\Sigma^{\infty}_{n=1} \frac{1}{5^{n}}$ is convergent.
Second
$\Sigma^{\infty}_{n=1} \frac{1}{n}$
$s_{1} =1$
$s_{2} =1 + \frac{1}{2}$
$s_{3}= 1 + \frac{1}{2} + (\frac{1}{3}+\frac{1}{4})$
$\gt 1+\frac{1}{2} + (\frac{1}{4}+\frac{1}{4})$
$=1+\frac{2}{2}$
$s_{4}=1 + \frac{1}{2} + (\frac{1}{3}+\frac{1}{4}) + (\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})$
$\gt 1 + \frac{1}{2} + (\frac{1}{4}+\frac{1}{4}) + (\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})$
$=1+\frac{1}{2} +\frac{1}{2} +\frac{1}{2}$
$=1+\frac{3}{2}$
$S_{2n} \gt 1+\frac{n}{2}$
Which shows that $S_{2n}ā\infty$ as $nā\infty$
Therefore the series $\Sigma^{\infty}_{n=1}\frac{1}{n}$ is divergent.
Since the first series is convergent and the second is divergent, the sum of the series is divergent.
