Question

How do you use the ratio test to test the convergence of the series `∑( (3n)!) / (n!)^3` from n=1 to infinity?

Answer 1

The series sum is divergent.

Explanation:

`((3n)!)/(n !)^3 =( (Pi_(k=1)^(n)k)(Pi_{k=n+1}^(2n)k)(Pi_(k=2n+1)^(3n)k))/((Pi_(k=1)^nk)(Pi_(k=1)^nk)(Pi_(k=1)^nk))`
`=((Pi_{k=n+1}^(2n)k)(Pi_(k=2n+1)^(3n)k))/((Pi_(k=1)^nk)(Pi_(k=1)^nk))`
`=Pi_{k=1}^n((k+n)/k)Pi_{k=1}^n((k+2n)/k)`
`=Pi_{k=1}^n(1+n/k)Pi_{k=1}^n(1+(2n)/k)`

Those are products of `2n` numbers whose value is greater than unity. So

`lim_{n->oo}((3n)!)/(n !)^3=oo` and the series sum is divergent.