Given:
The sequence is
an=1⋅3⋅5⋅...⋅(2n−1)(2n)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.
Calculation:
Obtain the first 20 terms of the sequence.
n 
1⋅3⋅5⋅...⋅(2n−1)

(2n)n

an=1⋅3⋅5⋅...⋅(2n−1)(2n)n

1 
1

21=2

121=0.5000

2 
3

42=16

342=0.1875

3 
15

63=216

1563=0.0694

4 
105

84=4,096

10584=0.0256

5 
945

105=100,000

945105=0.0095

6 
10,395

126=2,985,984

10,395126=0.0104

7 
135,135

147=105,413,504

135,135147=0.0013

8 
2,027,025

168=4,294,967,296

2,027,025168=0.0005

9 
34,459,425

189=198,359,290,368

34,459,4259!=0.0002

10 
654,729,075

2010=10,240,000,000,000

654,729,0752010=0.0001

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 convergent.
Also, the value of the limit is guessed to 0 as the plotted points are closer to 0.
To prove:
The sequence
an=1⋅3⋅5⋅...⋅(2n−1)(2n)n
is convergent to 0.
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)(2n)n
.
limn→∞1⋅3⋅5⋅...⋅(2n−1)(2n)n=limn→∞1⋅3⋅5⋅...⋅(2n−1)2n⋅2n⋅2n⋅...⋅2n︸n times=limn→∞(12n⋅32n⋅52n⋅...⋅2n−12n)=limn→∞(12n⋅32n⋅52n⋅...⋅(1−12n))
Use the limit property
limn→a[f(x)⋅g(x)]=limn→af(x)⋅limn→ag(x)
and simplify the terms as given below.
limn→∞1⋅3⋅5⋅...⋅(2n−1)(2n)n=limn→∞(12n)⋅limn→∞(32n)⋅limn→∞(52n)⋅...⋅limn→∞(1−12n)=1∞⋅3∞⋅5∞⋅...⋅(1−1∞)=0⋅0⋅0⋅...⋅(1)=0
Therefore,
limn→∞1⋅3⋅5⋅...⋅(2n−1)(2n)n=0
.
Since
limn→∞an
exists and by using the definition of the sequence, it can be concluded that the sequence is convergent to the limit 0.
Therefore, the sequence is convergent to the limit 0.
Hence the required proof is obtained.