a2=a1+1/a1=2+1/2=5/2
a3=a2+1/a2=2/5+5/2=29/10
数学归纳法证明
n=1时
a1=2>根号3,成立
假设n=k时成立
A(k)>√(2k+1)
令A(k)^2=(2k+1)+m
A(k+1)=(A(k)^2+1)/A(k)=[2(k+1)+1+(m-1)]/√[2(k+1)+1+(m-2)]
>[2(k+1)+1]/√[2(k+1)+1]
=√[2(k+1)+1]
n=k+1时也成立
设数列{an}满足a1=2,an+1=an+1/an,(n=1,2,3…).(1)证明：an＞（2n+1）1/2对一切正
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