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

1 Answer
Cesareo R.
Aug 25, 2016

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.

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