Question

已知{an}是公差不为零的等差数列，a1=1 ,a1，a3，a9成等比数列，设数列{1/an^2}的前n项和为Tn，证明Tn<7/4

Answer 1

核心方法：放缩与裂项相消法。1/n²<1/[(n-1)n]=1/(n-1) -1/n
设公差为d≠0,，因为a1，a3，a9成等比数列，所以
(a1+2d)²=a1(a1+8d)
即(1+2d)²=1+8d ，解得 d=1
所以 an=a1+(n-1)d=n
Tn=1+1/2²+1/3²+1/4²+…+1/n²
<1+1/4+1/(2×3)+ 1/(3×4)+…+1/[(n-1)n]
=1+1/4+1/2-1/3+1/3-1/4+…+1/(n-1)-1/n
=7/4 -1/n<7/4

Answer 2

a3=a1+2d
a9=a1+8d

(a3)^2=a1*a9
(a1+2d)^2=a1*(a1+8d)
(a1)^2+4a1d+4d^2=(a1)^2+8a1d
4d=4a1
a1=d=1
an=a1+(n-1)d
=1+n-1
=n

1/(an)^2
=1/n^2

1/n^2<1/[(n-1)n]=1/(n-1)-1/n

sn=1/1^2+1/2^2+..........+1/n^2
<1+1/(1*2)+1/(2*3)+1/(3*4)+……+1/[(n-1)*n]
=1+1-1/2+1/2-1/3+1/3-1/4+……+1/(n-1)-1/n
=2-1/n

发现放缩过大了，那么就在前面几项收一点。
原式从1/3^2以后开始放缩。
那么原式<1+1/2^2+1/2-1/3+1/3-1/4+…+1/(n-1)-1/n=7/4-1/n<7/4
所以Tn<7/4

Answer 3

设公差为d
a3^2=a1*a9 即(1+2d)^2=1*(1+8d)
解得d=1或0(舍去)
所以an=n
Tn=1/1^2+1/2^2+1/3^2……+1/n^2<1+1/(1*3)+1/(2*4)+……+1/(n-1)(n+1)
=1+1/2[(1-1/3)+(1/2-1/4)+……+(1/(n-1)-1/(n+1))]（裂项相消法）
=1+1/2(1+1/2-1/n-1/(n+1))<7/4