不等式和数列 (5 15:41:6)

题目

不等式和数列 (5 15:41:6)
已知正项数列｛an｝满足an2n-an-1   求证an

答案

仅供参考:
在正项数列中,an^2<=an-a(n+1),求证:an<1/n

a(n+1)<=an-an^2=an(1-an)
an>0,a(n+1)>0,所以an<1
a1<1,a2<=0.5*(1-0.5)=1/4
所以当n=2时an<=1/n+2成立
假设n>=2时有a(n)<=1/n+2<=1/4
因为x-x^2在[0,1/4]上为增函数
则a(n+1)<=a(n)(1-a(n))<=1/(n+2)(1-1/n+2))=(n+1)/(n^2+4n+4)<1/(n+2)
由数学归纳法可得,对任意n>=2都有an<=1/n+2
(当n>2时,等号不成立）
an<=1/n+2 <1/n