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, lim C(n)as n->infinity = square root (2pi)

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

Now if you can simplify the middle of the inequality so that it ends up being c, then you're done. But I dont know how...maybe you need to use a gamma function?

Let me try an identity here to simplify (2n)!
(2n)! = 1*3*5*7...* 2n-1 *2^n * n!

2/pi (1/(2n+1)) <= (1*3*5...*2n-1)^2 / (2^n * n!)^4 <= 1/2n *2/pi

now I square root.

square root (2/pi (1/(2n+1)) <= (1*3*5...*2n-1) / (2^n) *n!)^2 <= square root (1/2n *2/pi)

At this point I am stuck.
Hopefully someone has a clue?