009C Sample Midterm 2, Problem 2 Detailed Solution

Determine convergence or divergence:

$\sum _{n=1}^{\infty }{\frac {3^{n}}{n}}$ Background Information:
Direct Comparison Test
Let  $\{a_{n}\}$ and  $\{b_{n}\}$ be positive sequences where  $a_{n}\leq b_{n}$ for all  $n\geq N$ for some  $N\geq 1.$ 1. If  $\sum _{n=1}^{\infty }b_{n}$ converges, then  $\sum _{n=1}^{\infty }a_{n}$ converges.
2. If  $\sum _{n=1}^{\infty }a_{n}$ diverges, then  $\sum _{n=1}^{\infty }b_{n}$ diverges.

Solution:

Step 1:
First, we note that
${\frac {3^{n}}{n}}>0$ for all  $n\geq 1.$ This means that we can use a comparison test on this series.
Let  $a_{n}={\frac {3^{n}}{n}}.$ Step 2:
Let  $b_{n}={\frac {1}{n}}.$ We want to compare the series in this problem with
$\sum _{n=1}^{\infty }{\frac {1}{n}}.$ This is the harmonic series (or  $p$ -series with  $p=1.$ )
Hence,  $\sum _{n=1}^{\infty }b_{n}$ diverges.
Step 3:
Also, we have  $b_{n} since
${\frac {1}{n}}<{\frac {3^{n}}{n}}$ for all  $n\geq 1.$ Therefore, the series  $\sum _{n=1}^{\infty }a_{n}$ diverges
by the Direct Comparison Test.