009C Sample Midterm 1, Problem 4

Determine the convergence or divergence of the following series.

Be sure to justify your answers!

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

Foundations:
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 $a_{n}<b_{n}$ 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.

Final Answer:
converges

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