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.