009C Sample Midterm 1, Problem 4 Detailed Solution

Determine the convergence or divergence of the following series.

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