……不用楼上这么复杂吧Tn=1*2*...*n/((n+1)*(n+2)*...*(2n))=1/(n+1)*2/(n+2)*...*n/(2n)≤(1/2)^n故Tn→0(n→∞)
xn=(n！)²/(2n)！=n！/[(n+1)(n+2)…(2n)]=1/(n+1)·2/(n+2)·…·n/(2n)∴0＜xn＜1/(n+1)∵lim 1/(n+1)=0根据夹逼准则limxn=0
