009C Sample Midterm 1, Problem 4 Detailed Solution

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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty a_n}   converges
by the Direct Comparison Test.

Final Answer:
converges (by the Direct Comparison Test)

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