Calculus Problem about LIMIT ( Urgent )?

Using the Ratio Test:

r = lim(n→∞) [((n+1)!)^2/((2n+2)! + 2(n+1)!)] / [(n!)^2/((2n)! + 2(n!))]
..= lim(n→∞) [((n+1)!)^2/((2n+2)! + 2(n+1)!)] * [((2n)! + 2(n!))/(n!)^2]
..= lim(n→∞) (n+1)^2 * ((2n)! + 2(n!)) / ((2n+2)! + 2(n+1)!)
..= lim(n→∞) (n+1)^2 * ((2n)! + 2(n!)) / ((2n+2)(2n+1)(2n)! + 2(n+1)n!)
..= lim(n→∞) (n+1)^2 * ((2n)! + 2(n!)) / (2(n+1) ((2n+1)(2n)! + n!)
..= (1/2) lim(n→∞) (n+1) ((2n)! + 2(n!)) / ((2n+1)(2n)! + n!)
..= (1/2) lim(n→∞) ((n+1)/(2n+1)) * (1 + 2(n!)/(2n)!) / (1 + n!/((2n+1)(2n)!))

Noting that lim(n→∞) n!/(2n)! = 0 (via Squeeze Theorem), we have
r = (1/2) * 1/2 * (1 + 0) * (1 + 0)
..= 1/4.

Since r = 1/4 < 1, the series converges.

I hope this helps!