Given:
The sequence is
an=1⋅3⋅5⋅...⋅(2n−1)n!
.
Definition used:
If
an
is a sequence and
limn→∞an
exists, then the sequence
an
is said to be convergent; otherwise it is divergent.
Result used:
The sequence
{rn}
is divergent when
r>1
.
That is,
limn→∞rn=∞ if r>1
.
Calculation:
Obtain the first 10 terms of the sequence.
n 
1⋅3⋅5⋅...⋅(2n−1)

an=1⋅3⋅5⋅...⋅(2n−1)n!

1 
1=1

11!=1.0000

2 
1⋅3=3

32!=1.5000

3 
1⋅3⋅5=15

153!=2.5000

4 
1⋅3⋅5⋅7=105

1054!=4.3750

5 
1⋅3⋅5⋅7⋅9=945

9455!=7.8750

6 
1⋅3⋅5⋅7⋅9⋅11=10,395

10,3956!=14.4375

7 
1⋅3⋅5⋅...⋅11⋅13=135,135

135,1357!=26.8125

8 
1⋅3⋅5⋅...⋅13⋅15=2,027,025

2,027,0258!=50.2734

9 
1⋅3⋅5⋅...⋅15⋅17=34,459,425

34,459,4259!=94.9609

10 
1⋅3⋅5⋅...⋅17⋅19=654,729,075

654,729,07510!=180.4258

Plot the points
(n,an), for n=1,2,...10
on the graph as shown below in Figure 1.
From the graph, it is observed that the sequence is divergent because the sequence is increase as large value of n.
To prove:
The sequence
an=1⋅3⋅5⋅...⋅(2n−1)n!
diverges.
Proof:
Obtain the limit of the sequence (the value of the term
an
as n tends to infinity).
That is, compute
limn→∞an=limn→∞1⋅3⋅5⋅...⋅(2n−1)n!
. (1)
Here, it is observed that the numerator
(1⋅3⋅5⋅...⋅(2n−1))
will increase more faster than the denominator
(n!)
.
Claim: To prove by induction that
an≥(32)n−1 for all n
.
Base case:
n=1
a1≥(32)1−111!≥(32)01≥1
Thus, the claim is true for
n=1
.
Induction hypothesis:
n=k
Assume that the claim is true when
n=k
.
That is,
ak≥(32)k−1
is true for all k.
Inductive step:
n=k+1
To prove that the claim is true when
n=k+1
.
ak+1=1⋅3⋅5⋅...⋅(2k−1)⋅(2(k+1)−1)(k+1)!=1⋅3⋅5⋅...⋅(2k−1)⋅(2k+1)(k+1)⋅k!=1⋅3⋅5⋅...⋅(2k−1)k!⋅2k+1k+1=ak⋅2k+1k+1
Use the induction hypothesis
ak≥(32)k−1 for all k
.
ak+1≥(32)k−1⋅2k+1k+1
Since
2k+1k+1≥32
,
2(2k+1)≥3(k+1)⇔4k+2≥3k+3⇔k≥1
.
ak+1≥(32)k−1⋅32=(32)k−1+1=(32)k
Therefore, by induction, the claim
an≥(32)n−1
is true for all n.
From equation (1),
limn→∞an≥limn→∞(32)n−1
.
Since
32
is greater than one and by the result of divergent sequence, the term
(32)n−1
diverges.
Therefore,
limn→∞an=∞
.
Since
limn→∞an
does not exist and by the definition of the sequence, it can be concluded that the sequence is divergent.
Therefore, the sequence is divergent.