数学表达式n!=

1*2*3*4*......*n
n!=1*2*3*...*n
n1!=n*(n-1)*(n-2)*(n-3).......3*2*1

你是不是要斯特林公式呀 一下证明
令a(n)=n! / [ n^(n+1/2) * e^(-n) ]
则a(n) / a(n+1) = (n+1)^(n+3/2) / [ n^(n+1/2) * (n+1) * e ]
=(n+1)^(n+1/2) / [ n^(n+1/2) * e]
=(1+1/n)^n * (1+1/n)^1/2 *1/e
当n→∞时,(1+1/n)^n→e,(1+1/n)^1/2→1
即lim(n→∞) a(n)/a(n+1)=1
所以lim(n→∞)a(n) 存在
设A=lim(n→∞)a(n)
A=lim(n→∞)n! / [ n^(n+1/2) * e^(-n) ]
利用Wallis公式,π/2 = lim(n→∞)[ (2n)!! / (2n-1)!! ]^2 / (2n+1)
π/2 = lim(n→∞)[ (2n)!! / (2n-1)!! ]^2 / (2n+1)
=lim(n→∞)[ (2n)!! * (2n)!! / (2n)! ]^2 / (2n+1)
=lim(n→∞) 2^(4n) [ (n!)^2 / (2n)! ]^2 / (2n+1)
=lim(n→∞) 2^(4n) [ (A * n^(n+1/2) * e^(-n) )^2 / (A * (2n)^(2n+1/2) * e^(-2n) )]^2 / (2n+1)
=lim(n→∞) 2^(4n) [ 2^(-2n-1/2) * A * √n ]^2 / (2n+1)
=lim(n→∞) 2^(4n) * A^2 * 2^(-4n-1) * n/(2n+1)
=A^2 / 4
所以A=√(2π)
lim(n→∞)n! / [ n^(n+1/2) * e^(-n) ] = √(2π)
即lim(n→∞) √(2πn)