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
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 \frac{n^3+7n}{\sqrt{1+n^{10}}}>0}
for all  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 n\ge 1.}
This means that we can use a comparison test on this series.
Let  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 a_n=\frac{n^3+7n}{\sqrt{1+n^{10}}}.}

Step 2:
Let  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 b_n=\frac{1}{n^2}.}
We want to compare the series in this problem with
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 b_n=\sum_{n=1}^\infty \frac{1}{n^2}.}
This is a  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 p} -series with  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 p=2.}
Hence,  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 b_n}   converges

Step 3:
Now, we have
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 \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
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{n^3+7n}{\sqrt{1+n^{10}}}}
converges by the Limit Comparison Test.

Final Answer:
converges

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