已知等差数列{an}的前n项和为Sn,bn=1/sn,且a3b3=1/2,S3+S5=21,

题目

已知等差数列{an}的前n项和为Sn,bn=1/sn,且a3b3=1/2,S3+S5=21,
（1）求数列{bn}的通项公式；
（2）求证：b1+b2+…+bn〈 2
要求有适当的解析说明,越清楚越好,

答案

设等差数列{a(n)}的通项为
a(n) = c + (n-1)d,n = 1,2,...
前n项和为S(n) = nc + n(n-1)d/2,n = 1,2,...
b(n) = 1/s(n) = 1/[nc + n(n-1)d/2],
又 a(3)b(3) = 1/2,S(3) + S(5) = 21,
(c + 2d)/[3c + 3d] = 1/2,
3c + 3d + 5c + 10d = 21,
c = d = 1.
（1）数列{b(n)}的通项公式为,
b(n) = 1/[n + n(n-1)/2] = 2/[n(n+1)],n=1,2,...
（2）b(1) + b(2) +…+ b(n)
= [2/2] + [2/6] + ...+ 2/[n(n+1)]
= 2{1/2 + 1/6 + ...+ 1/[n(n+1)]}
= 2{[1 - 1/2] + [1/2 - 1/3] + ...+ [1/n - 1/(n+1)]}
= 2[1 - 1/(n+1)]
< 2.