设正数列{an}满足an2

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设正数列{an}满足an2
设正数列{an}满足an2<=an-an+1,证明an<=1/n+2(n=2,3,3,...)

设正数列{an}满足an2
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)

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