1. Use the following derived inequality (i used integrals of sin ^2n tdt 0 to pi/2 and sin ^2n+1 tdt 0 to pi/2, i can post the derivation if you request):

((2^n * n! )^2 ) / ((2n)!*(2n+1)) <= pi/2 * ((2n)! / (2^n * n! )^2) <= ((2^n * n! )^2) / ((2n)!* (2n))

(Be careful with the squaring.for example (2^n * n! )^2 = (2^2n) n!n!)

2. Substitute c*n^(n+1/2)* e^(-n) for n! in the inequality,
3. Show that the limit C of the sequence C(n) as n goes to infinity as defined above satisfies the following inequality:
square root (2pi) <= C <= square root ((2pi (1+1/2n))
(in other words,show lim C(n)as n->infinity = square root (2pi)