009C Sample Final 3, Problem 4

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

(a)  $\sum _{n=1}^{\infty }{\frac {n!}{(2n)!}}$

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

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

Solution:

(a)

Step 1:
We begin by using the Ratio Test.
We have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {a_{n+1}}{a_{n}}}{\bigg |}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)!}{(2(n+1))!}}{\frac {(2n)!}{n!}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg |}{\frac {(n+1)n!}{(2n+2)(2n+1)(2n)!}}{\frac {(2n)!}{n!}}{\bigg |}}\\&&\\&=&\displaystyle {\lim _{n\rightarrow \infty }{\frac {n+1}{(2n+2)(2n+1)}}}\\&&\\&=&\displaystyle {0.}\end{array}}$

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

(b)

Step 1:
For
$\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{n+1}},$
we notice that this series is alternating.
Let  $b_{n}={\frac {1}{n+1}}.$
First, we have
${\frac {1}{n+1}}\geq 0$
for all  $n\geq 1.$
The sequence  $\{b_{n}\}$  is decreasing since
${\frac {1}{n+2}}<{\frac {1}{n+1}}$
for all  $n\geq 1.$

Step 2:
Also,
$\lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {1}{n+1}}=0.$
Therefore,
$\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{n+1}}$
converges by the Alternating Series Test.

Final Answer:
(a)    converges (by the Ratio Test)
(b)    converges (by the Alternating Series Test)

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