an=4/n,Tn=a1^2+a2^2+a3^2+...+an^2,Sn=32-16/n,比较Tn与Sn的大小

Question

an=4/n,Tn=a1^2+a2^2+a3^2+...+an^2,Sn=32-16/n,比较Tn与Sn的大小

Answer 1

解,
a1 = 4 a2 = 2
T1 = 16 T2 = 20
S1 = 16 S2= 24
设当n = k >=2时,Sn > Tn,则
Sn+1 - Tn+1 = 32 - 16/(n+1) - Tn - (an+1)^2
= Sn - Tn + 16/n -16/(n+1) - (an+1)^2
> 16/n -16/(n+1) - (an+1)^2
= 16/n -16/(n+1) - 16/(n+1)^2
= [16/n/(n+1) ^2] [ (n+1)^2 - n(n+1)-n]
= 16/n/(n+1) ^2 >0
因此,知 Sn > =Tn,仅当 n = 1时取等.