an=(1+2+.+n)/n
an=[n(1+n)/2]/n=(1+n)/2
a1=(1+1)/2=1
a(n+1)=(2+n)/2
1/an=2/(1+n)
1/a(n+1)=2/(2+n)
bn=1/[ana(n+1)]
=[2/(1+n)][2/(2+n)]
=4/[(1+n)(2+n)]
=4[1/(1+n)-1/(2+n)]求得an=1/2n+1/2
an=（n+1）/2
bn=4/(n+1)(n+2)=4(1/(n+1)-1/(n+2))
bn的前n項(xiàng)和為4*（1/2-1/3+1/3-1/4+……+1/(n+1)-1/(n+2)）=4(1/2-1/(n+2))=2n/(n+2)