# 009C Sample Final 3, Problem 3

Test if the following series converges or diverges. Give reasons and clearly state if you are using any standard test.

${\displaystyle \sum _{n=1}^{\infty }{\frac {n^{3}+7n}{\sqrt {1+n^{10}}}}}$
Foundations:
Limit Comparison Test
Let  ${\displaystyle \{a_{n}\}}$  and  ${\displaystyle \{b_{n}\}}$  be positive sequences.
If  ${\displaystyle \lim _{n\rightarrow \infty }{\frac {a_{n}}{b_{n}}}=L,}$  where  ${\displaystyle L}$  is a positive real number,
then  ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$  and  ${\displaystyle \sum _{n=1}^{\infty }b_{n}}$  either both converge or both diverge.

Solution:

Step 1:
First, we note that
${\displaystyle {\frac {n^{3}+7n}{\sqrt {1+n^{10}}}}>0}$
for all  ${\displaystyle n\geq 1.}$
This means that we can use a comparison test on this series.
Let  ${\displaystyle a_{n}={\frac {n^{3}+7n}{\sqrt {1+n^{10}}}}.}$
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:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {a_{n}}{b_{n}}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {({\frac {n^{3}+7n}{\sqrt {1+n^{10}}}})}{({\frac {1}{n^{2}}})}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n^{5}+7n^{3}}{\sqrt {1+n^{10}}}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n^{5}+7n^{3}}{\sqrt {1+n^{10}}}}{\bigg (}{\frac {\frac {1}{n^{5}}}{\frac {1}{n^{5}}}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {1+{\frac {7}{n^{4}}}}{\sqrt {{\frac {1}{n^{10}}}+1}}}}\\&&\\&=&\displaystyle {1.}\end{array}}}$
Therefore, the series
${\displaystyle \sum _{n=1}^{\infty }{\frac {n^{3}+7n}{\sqrt {1+n^{10}}}}}$
converges by the Limit Comparison Test.