X1=a>0,Y1=b>0,Xn+1=(Xn+Yn)/2,Yn+1=(Xn*Yn)^1/2,求证数列Xn,Yn的极限相等.其中两个n+1均为下角标

首先证极限的存在性根据不等式性质,X(n+1)≥Y(n+1) (对于任意n≥1）,所以 X(n+2)=(X(n+1)+Y(n+1))/2≤X(n+1),Y(n+2)=(X(n+1)*Y(n+1))^1/2≥Y(n+1).所以任意n>2Y2≤Y3≤...≤Y(n-1)≤Yn≤Xn≤Xn-1≤...≤X3≤X2所...