What is the conclusion of the ratio test when applied to Σ,∞,n=1 (n!)^2/(3n)!

Look at the ratio of successive terms:

((n+1)!²/(3n+3)!)/(n!²/(3n)!)

=((n+1)!²/n!²)/((3n+3)!/(3n)!)

=(n+1)²/((3n+3)(3n+2)(3n+1))

The highest order terms in the top and bottom are n²/(27n³)

In the limit as n→∞, the ratio goes to zero, so the series converges.