因为1+2+3+.+n=n(n+1)/2
所以1/(1+2+3+...+n)=2/n(n+1)=2[(1/n)-(1/n+1)]
原式=2[(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/n-1/n+1)]
=2[1-(1/n+1)]
=2n/n+1