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