A(n+1)=An+1/An
两端平方得
A(n+1)^2=An^2+1/An^2+2
A(n+1)^2-An^2=1/An^2+2>2
从1到n-1相加,得到
An^2-A1^2>2(n-1)
An^2>2(n-1)+4>2n+1,开方即可.
把An=Bn*根号n 代入A(n+1)=An+1/An整理得到
(Bn+1-Bn)*根号(n+1)=1/(Bn*根号n)+Bn/(根号n+根号(n+1))
=1/An-An/【(根号n+根号(n+1))根号n】
=(1/An)(1-An^2/【(根号n+根号(n+1))根号n】)