# 009C Sample Midterm 3, Problem 3 Detailed Solution

Test if each the following series converges or diverges.

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

(a)  $\sum _{n=1}^{\infty }}\,{\frac {n!}{(3n+1)!}}$ (b)  $\sum _{n=2}^{\infty }}\,{\frac {\sqrt {n}}{n^{2}-3}}$ Background Information:
1. Ratio Test
Let  $\sum a_{n}$ be a series and  $L=\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}.$ Then,

If  $L<1,$ the series is absolutely convergent.

If  $L>1,$ the series is divergent.

If  $L=1,$ the test is inconclusive.

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

(a)

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

${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)!}{(3(n+1)+1)!}}{\frac {(3n+1)!}{n!}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {(n+1)n!}{(3n+4)!}}{\frac {(3n+1)!}{n!}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {(n+1)(3n+1)!}{(3n+4)(3n+3)(3n+2)(3n+1)!}}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n+1}{(3n+4)(3n+3)(3n+2)}}}\\&&\\&=&\displaystyle {0.}\end{array}}$ Step 2:
Since  $0<1,$ the series is absolutely convergent by the Ratio Test.
Therefore, the series converges.

(b)

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