显然通项公式为an=1/(1+2+3+.......n)=1/[n(n+1)/2]=2/[n(n+1)]=2[1/n-1/(n+1)],本题就是求数列{an}的前n项和，所以Sn=2[1-1/2+1/2-1/3+……+1/n-1/(n+1)]=2/[1-1/(n+1)]=2n/(n+1)
