设a[n]=(n!)^2/(2n)!=n!/((n+1)(n+2)...(2n)),
因为a[n+1]/a[n]=(n+1)/(4x+2)0,所以a[n]有极限,设为A,
所以a[n+1]=a[n]*(n+1)/(2(2n+1))=a[n]*(1+1/n)/(4+2/n)
令n→∞得
A=A*1/4
A=0