阶乘.求证:1/(1+1)!+2/(2+1)!+…+n/(n+1)=1-1/(n+1)!求证：1/(1+1)!+2/(2+1)!+…+n/(n+1)=1-1/(n+1)!

题目应该是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
证毕。

证明：
1.当n=1时，左边=1/(1+1)!=1/2
右边=1-1/(1+1)!=1/2=左边
2.假设n=k时，1/(1+1)!+2/(2+1)!+…+k/(k+1)!=1-1/(k+1)!
那么n=k+1时，
左边=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时等式也成立
所以原命题得证

用数学归纳法.
（1）当n=1时,1/(1+1)!=1/2=1-1/(1+1)!
（2）假设当n=k时等式成立,即1/(1+1)!+2/(2+1)!+…+k/(k+1)!=1-1/(k+1)!
那么,当n=k+1时,
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)!
所以当n=k+1时,原等式依然成立.
综合（1）（2）可得,原等式成立.