1/(1*2)+2/(1*2*3)+3/(1*2*3*4)+...+n/(1*2*3*...*(n+1))

来源：百度知道 编辑：UC知道 时间：2024/06/14 12:39:11
1/(1*2)+2/(1*2*3)+3/(1*2*3*4)+...+n/(1*2*3*...*(n+1))

1/(1*2)+2/(1*2*3)+3/(1*2*3*4)+...+n/(1*2*3*...*(n+1))
=1-1/2+1/2-1/(2*3)+1/(2*3)-1/(2*3*4)+...+1/(2*3*..*n)-1/(3*4*5*....*(n+1))
=1-1/(3*4*5*....*(n+1))

=1/(1*2)+1/(1*2)-1/(1*2*3)+1/(1*2*3)......-1/(1*2*3*...*(n+1))=1-1/(1*2*3*...*(n+1))

n/(1*2*3*...*(n+1)=[1/n!-1/(n+1)!]
所以原式＝1-1/(n+1)!

取an＝n/(1*2*3*...*(n+1))，则
an=1/(1*2*3*...*n)-1/(1*2*3*...*(n+1))
原式＝a1+a2+a3+...+an
=1/1-1/1*2+1/1*2-1/1*2*3+1/1*2*3-1/1*2*3*4+...+1/(1*2*3*...*n)-1/(1*2*3*...*(n+1))
=1-1/(1*2*3*...*(n+1)).

an=1/(1*2*3*...*n)-1/(1*2*3*...*(n+1))
原式＝a1+a2+a3+...+an
=1/1-1/1*2+1/1*2-1/1*2*3+1/1*2*3-1/1*2*3*4+...+1/(1*2*3*...*n)-1/(1*2*3*...*(n+1))
=1-1/(1*2*3*...*(n+1)).