已知各项为正数的数列{an}的前n项和为{Sn},首项为a1=1,点p（根号Sn,Sn+1）在曲线

a(n)>0.
s(n) >0.
s(n+1) = {[s(n)]^(1/2) + 1}^2,
[s(n+1)]^(1/2) = [s(n)]^(1/2) + 1,
{[s(n)]^(1/2)}是首项为[s(1)]^(1/2) = [a(1)]^(1/2) = 1,公差为1的等差数列.
[s(n)]^(1/2) = 1 + (n-1) = n,
s(n) = n^2.
a(n+1) = s(n+1)-s(n) = (n+1)^2 - n^2 = 2n+1.
a(n) = 2(n-1) + 1= 2n-1.
a(2) = 2*2-1 = 3,
a(3) = 2*3-1 = 5.
b(n) = 1/[a(n)a(n+1)] = 1/[(2n-1)(2n+1)] = (1/2)[1/(2n-1) - 1/(2n+1)],
t(n) = b(1) + b(2) + ... + b(n-1) + b(n)
= (1/2)[1/1 - 1/3 + 1/3 - 1/5 + ... + 1/(2n-3) - 1/(2n-1) + 1/(2n-1) - 1/(2n+1)]
= (1/2)[1 - 1/(2n+1)]
= n/(2n+1)
= 1/(2 + 1/n)
1=1/n>0,
3>= 2 + 1/n > 2,
1/3