009C Sample Final 3, Problem 3 Detailed Solution

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

$\sum _{n=1}^{\infty }{\frac {n^{3}+7n}{\sqrt {1+n^{10}}}}$

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Background Information:
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.

Final Answer:
converges (by the Limit Comparison Test)

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