已知等差数列｛an｝的前项和为Sn,bn=1/Sn,且a3b3=1/2,S3+S5=21
问题描述：
已知等差数列｛an｝的前项和为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
便于我理解!
答
an=a1+d(n-1)
=>
Sn=(a1+an)n/2
=(2a1+d(n-1))n/2
=>
bn=2/(n(2a1+d(n-1)))
=>
b3=2/(3(2a1+2d))=1/(3a1+3d)
=>
a3/(3a1+3d)=1/2
=>
(a1+2d)/(3a1+3d)=1/2
S3=(2a1+2d)3/2=3a1+3d
S5=(2a1+4d)5/2=5a1+10d
=>
3a1+3d+5a1+10d=21
=>
8a1+13d=21
结合(a1+2d)/(3a1+3d)=1/2
得到a1=1,d=1
=>
an=n
Sn=(1+n)n/2
bn=1/Sn=2/(n^2+n)=2/n-2/(n+1)
b1+b2+...bn
=2/1-2/2+2/2-2/3+2/3-2/4.+2/n-2/(n+1)
=2-2/(n+1)