1/2!+2/3!+3/4!+……+n/（n+1）!
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1/2!+2/3!+3/4!+……+n/（n+1）!1/2!+2/3!+3/4!+……+n/（n+1）!1/2!+2/3!+3/4!+……+n/（n+1）!套公式：n/(n+1)!=1/n!-1/(n
1/2!+2/3!+3/4!+……+n/（n+1）!
1/2!+2/3!+3/4!+……+n/（n+1）!
1/2!+2/3!+3/4!+……+n/（n+1）!
套公式：n/(n+1)!=1/n!-1/(n+1)!
原式=1-1/2!+1/2!-1/3!+…+1/n!-1/(n+1)!
=1-1/(n+1)!
...........1/(n+1)*(n-1)!
这样做：
原式=1/2!+2/3!+3/4!+……+n/（n+1）！
=(1/2!+1/2!)+(2/3!+1/3!)+(3/4!+1/4!)+......+(n/(n+1)!+1/(n+1)!)-(1/2!+1/3!+1/4!+......+1/n!+1/(n+1)!)
=(1+3/3!+4/4!+......+n/n!+(n+1)/(n+1)!)-(1...
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这样做：
原式=1/2!+2/3!+3/4!+……+n/（n+1）！
=(1/2!+1/2!)+(2/3!+1/3!)+(3/4!+1/4!)+......+(n/(n+1)!+1/(n+1)!)-(1/2!+1/3!+1/4!+......+1/n!+1/(n+1)!)
=(1+3/3!+4/4!+......+n/n!+(n+1)/(n+1)!)-(1/2!+1/3!+1/4!+......+1/n!+1/(n+1)!)
=(1+1/2!+1/3!+1/4!+......+1/n!)-(1/2!+1/3!+1/4!+......+1/n!+1/(n+1)!)
=1-1/(n+1)!
思路就是，在原式每项的后边加上相应阶乘的倒数，然后再在最后减去相应阶乘的倒数，会发现两两抵消，到最后只剩下排头的1和最后的1/(n+1)!
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