2Sn=an^2+an
2(Sn-Sn-1)=an^2+an-(an-1^2+an-1)
2an=an^2+an-(an-1^2+an-1)
an^2-an-an-1^2-an-1=0
因式分解
an^2-an-1^2-(an+an-1)=0
(an+an-1)(an-an-1-1)=0
因为{an}为正数数列
只能an-an-1-1=0
an-an-1=1,是等差数列
2S1=2a1=a1^2+a1
a1(a1-1)=0
a1=1
通项公式an=1+(n-1)*1=n
Tn=1+1/2^2+1/3^3+...+1/n^2
n>=3时
Tn>1+1/(2*3)+1/(3*4)+...+1/n(n+1)
=1+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)
=3/2-1/(n+1)
又因为n>=3
3/2-1/(n+1)-[3/2+(1-2n)/2n^2]
=-1/(n+1)-1/2n^2+1/n
=1/n-1/(n+1)-1/2n^2
=1/n(n+1)-1/2n^2
=1/(n^2+n)-1/(n^2+n^2)
>0
所以
Tn>3/2-1/(n+1)>3/2+(1-2n)/2n^2
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