k階乘分之一求和公式推導過程？

用數學歸納法.

（1）當n=1時,1/(1+1)!=1/2=1-1/(1+1)!

（2）假設當n=k時等式成立,即1/(1+1)!+2/(2+1)!+…+k/(k+1)!=1-1/(k+1)!

那么,當n=k+1時,

1/(1+1)!+2/(2+1)!+…+k(k+1)!+(k+1)/(k+2)!

=1-1/(k+1)!+(k+1)/(k+2)!

=1-(k+2)/(k+2)!+(k+1)/(k+2)!

=1-1/(k+2)!

所以當n=k+1時,原等式依然成立.

綜合（1）（2）可得,原等式成立.

證明：

1.當n=1時，左邊=1/(1+1)!=1/2

右邊=1-1/(1+1)!=1/2=左邊

2.假設n=k時，1/(1+1)!+2/(2+1)!+…+k/(k+1)!=1-1/(k+1)!

那么n=k+1時，

左邊=1/(1+1)!+2/(2+1)!+…+k/(k+1)!+(k+1)/(k+1+1)!

=1-1/(k+1)!+(k+1)/(k+1)!(k+2)

=1-[1-(k+1)/(k+2)]/(k+1)!

=1-1/(k+2)(k+1)!

=1-1/(k+2)!

即n=k+1時等式也成立

題目應該是1/(1+1)!+2/(2+1)!+…+n/(n+1)！=1-1/(n+1)!吧？

左式+1/(n+1)!

= 1/(1+1)!+2/(2+1)!+…+n/(n+1)！+1/(n+1)!

=1/(1+1)!+2/(2+1)!+…+(n+1)/(n+1)!

=1/(1+1)!+2/(2+1)!+…+(n-1)/(n-1+1)!+1/n!

=1/(1+1)!+2/(2+1)!+…+(n-1)/n!+1/n!

=...

=1/(1+1)!+1/(1+1)!=1

證畢。