已知等差数列｛an}的前n项通项公式为Sn,且a1+a3=10,S4=24令Tn==1/S1+……1/Sn,求证Tn＜3/4

Question

已知等差数列｛an}的前n项通项公式为Sn,且a1+a3=10,S4=24 令Tn==1/S1+……1/Sn,求证Tn＜3/4

Answer 1

{an}的公差为d
a1+a3=a1+a1+2d=10
得到：a1+d=5 ①
S4=4(a1+a4)/2=2(a1+a1+3d)=24
得到：2a1+3d=12 ②
综合①② 得到：a1=3 d=2
Sn=na1+n(n-1)d/2=n(n+2)
1/Sn=1/n(n+2)=½[1/n-1/(n+2)] 这是关键
Tn=1/S1+……1/Sn
Tn=½[1/1-1/3+1/2-1/4+1/3-1/5+.+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=½[1+1/2-1/(n+1)-1/(n+2)]
＜3/4