# 009C Sample Midterm 3, Problem 3 Detailed Solution

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Test if each the following series converges or diverges.

Give reasons and clearly state if you are using any standard test.

(a)  ${\displaystyle {\displaystyle \sum _{n=1}^{\infty }}\,{\frac {n!}{(3n+1)!}}}$

(b)  ${\displaystyle {\displaystyle \sum _{n=2}^{\infty }}\,{\frac {\sqrt {n}}{n^{2}-3}}}$

Background Information:
1. Ratio Test
Let  ${\displaystyle \sum a_{n}}$  be a series and  ${\displaystyle L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}.}$
Then,

If  ${\displaystyle L<1,}$  the series is absolutely convergent.

If  ${\displaystyle L>1,}$  the series is divergent.

If  ${\displaystyle L=1,}$  the test is inconclusive.

2. If a series absolutely converges, then it also converges.
3. 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:

(a)

Step 1:
We begin by using the Ratio Test.
We have

${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)!}{(3(n+1)+1)!}}{\frac {(3n+1)!}{n!}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {(n+1)n!}{(3n+4)!}}{\frac {(3n+1)!}{n!}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {(n+1)(3n+1)!}{(3n+4)(3n+3)(3n+2)(3n+1)!}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n+1}{(3n+4)(3n+3)(3n+2)}}}\\&&\\&=&\displaystyle {0.}\end{array}}}$

Step 2:
Since  ${\displaystyle 0<1,}$  the series is absolutely convergent by the Ratio Test.
Therefore, the series converges.

(b)

Step 1:
First, we note that
${\displaystyle {\frac {\sqrt {n}}{n^{2}-3}}>0}$
for all  ${\displaystyle n\geq 2.}$
This means that we can use a comparison test on this series.
Let  ${\displaystyle a_{n}={\frac {\sqrt {n}}{n^{2}-3}}.}$
Step 2:
Let  ${\displaystyle b_{n}={\frac {\sqrt {n}}{n^{2}}}={\frac {1}{n^{\frac {3}{2}}}}.}$
We want to compare the series in this problem with
${\displaystyle \sum _{n=1}^{\infty }b_{n}=\sum _{n=2}^{\infty }{\frac {1}{n^{\frac {3}{2}}}}.}$
This is a  ${\displaystyle p}$-series with  ${\displaystyle p={\frac {3}{2}}.}$
Hence,  ${\displaystyle \sum _{n=2}^{\infty }b_{n}}$  converges.
Step 3:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {a_{n}}{b_{n}}}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {{\big (}{\frac {\sqrt {n}}{n^{2}-3}}{\big )}}{{\bigg (}{\frac {1}{n^{\frac {3}{2}}}}{\bigg )}}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {({\sqrt {n}})n^{\frac {3}{2}}}{n^{2}-3}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n^{2}}{n^{2}-3}}}\\&&\\&=&\displaystyle {1.}\end{array}}}$
Therefore, the series
${\displaystyle \sum _{n=2}^{\infty }{\frac {\sqrt {n}}{n^{2}-3}}}$
converges by the Limit Comparison Test.