# 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.

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

Solution:

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