数列An=1/n+2/n+3/n+……+n/n
问题描述:
数列An=1/n+2/n+3/n+……+n/n
Bn=2/An(An+1),(An+1表示An的后一项),
求Sn=B1+B2+……=Bn
答
事实上,An=(n+1)/2.这样,Bn=8/[(n+1)(n+2)]=8[1/(n+1)-1/(n+2)].根据此式,Sn=8[1/2-1/3+1/3-1/4+1/4-1/5+…+1/(n+1)-1/(n+2)]
=4-8/(n+2)