Question

已知等差数列an的前n项和为Sn,bn=1/sn 且 a3b3=1/2,s3+s5=21 求证b1+b2+.......+bn<2

Answer 1

由 a3b3=1/2 a3/S3=1/2 a3/(a1+a3)=1/2 a3=3a1
由 s3+s5=21 1/2(a1+a3)*3+1/2*2a3*5=21 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)<2

Answer 2

因为a3b3=1/2
b3=1/s3
所以a3/S3=1/2＝a3/(a1+a2+a3)=(a1+2d)/(3a1+3d)
a1=d
S3+S5=3a1+3d+(5a1+10d)=8a1+13d=21
a1=d=1

Sn=na1+n(n-1)d/2=n(n+1)/2
bn=2/n(n+1)=2[1/n-1/(n+1)]
b1+b2+b3+……+bn=2[(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/n-1/(n+1))]=2[1-1/(n+1)]
=2-2/(n+1)<2