Given information : The n^thterm of the sequence is given by, an=1  .  3  .  5  .  .  .  .  .  (2n − 1)(2n)n .….(1)

By substituting the value of n in equation (1), gives the first five terms as

  a1=1(2×1)1=12

  a2=1×3(2×2)2=316

  a3=1×3×5(2×3)3=572

  a4=1×3×5×7(2×4)4=1054096

  a5=1×3×5×7×9(2×5)5=18920,000

Table formed with the obtained values is shown below.

The graph of sequence is obtained as shown below, with n in x - axis and a_n in y - axis .

n	an
1	12
2	316
3	572
4	1054096
5	18920000

Above graph shows that the sequence is convergent sequence. And its limit is 0.

Proof of limit of a_n ,

  an=1  .  3  .  5  .  .  .  .  .  (2n − 1)(2n)n=(2n)!2n×n!(2n)n

  =(2n)!2n×n!×(2n)n

Therefore,

  limn→∞an=limn→∞(2n)!2n×n!×(2n)n

  =limn→∞12n×n!×limn→∞(2n)!(2n)n

  =0×0=0

Thus,

The obtained sequence is convergent and it’s limit is zero.