How do I compute this ratio series test

infΣn=1 ( (3n)!/(n!)^3))

Please show your working, I'm finding this one tough.

Thanks

James

r = lim(n→∞) |a(n+1) / a(n)|
= lim(n→∞) |[(3(n+1))! / ((n+1)!)^3] / [(3n)! / (n!)^3]|
= lim(n→∞) [(3n+3)! / ((n+1)!)^3] * [(n!)^3 / (3n)!]
= lim(n→∞) [(3n+3)(3n+2)(3n+1)(3n)! / ((n+1) n!)^3] * [(n!)^3 / (3n)!]
= lim(n→∞) (3n+3)(3n+2)(3n+1) / (n+1)^3
= 27/1, by comparing leading coefficients (both num. and denom. are of degree 3)
= 27.

Since r = 27 > 1, this series diverges by the Ratio Test.

I hope this helps!