009C Sample Midterm 1, Problem 4

Determine the convergence or divergence of the following series.

Be sure to justify your answers!

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 \frac{1}{n^23^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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