# 009C Sample Final 2, Problem 3

Determine if the following series converges or diverges. Please give your reason(s).

(a)  ${\displaystyle \sum _{n=0}^{\infty }{\frac {n!}{(2n)!}}}$

(b)  ${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{n+1}}}$

Foundations:
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. Alternating Series Test
Let  ${\displaystyle \{a_{n}\}}$  be a positive, decreasing sequence where  ${\displaystyle \lim _{n\rightarrow \infty }a_{n}=0.}$
Then,  ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}a_{n}}$  and  ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n+1}a_{n}}$
converge.

Solution:

(a)

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

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

Step 2:
Since
${\displaystyle \lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}=0<1,}$
the series is absolutely convergent by the Ratio Test.
Therefore, the series converges.

(b)

Step 1:
For
${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{n+1}},}$
we notice that this series is alternating.
Let  ${\displaystyle b_{n}={\frac {1}{n+1}}.}$
The sequence  ${\displaystyle \{b_{n}\}}$  is decreasing since
${\displaystyle {\frac {1}{n+2}}<{\frac {1}{n+1}}}$
for all  ${\displaystyle n\geq 0.}$
Step 2:
Also,
${\displaystyle \lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{n+1}}=0.}$
Therefore, the series  ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{n+1}}}$   converges
by the Alternating Series Test.