# 009C Sample Midterm 1, Problem 4 Detailed Solution

Determine the convergence or divergence of the following series.

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}3^{n}}}}$

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

Solution:

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