因为1/(1*2*3)=(1/2)*[1/(1*2)-1/(2*3)],
1/(2*3*4)=(1/2)*[1/(2*3)-1/(3*4)],
...
（可以把右边通分,证明等式成立）
所以1/(1*2*3)+1/(2*3*4)+...+1/n(n+1)(n+2)
=(1/2)*[1/(1*2)-1/(2*3)+1/(2*3)-1/(3*4)+...+1/n(n+1)-1/(n+1)(n+2)]
=(1/2)*[1/2-1/(n+1)(n+2)]
=1/4-1/2(n+1)(n+2)
因为1/2(n+1)(n+2)>0,所以式子左边=1/4-1/2(n+1)(n+2)