裂项相消法,错位相减法,倒序相加法,分类求和法,累加/累乘法

我给你举个例子：
求Sn=1/1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+……+n)
先求通项公式an=1/(1+2+3+……+n)=1/[n(n+1)/2]=2/n(n+1)
所以Sn=2[1/(1*2)+1/(2*3)+……+1/n(n+1)]
再裂项相消：Sn=2[(1-1/2)+(1/2-1/3)+……+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)