1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+1/(1+2+3+4+5)+......=?

数列通项an=1/[n(n+1)/2] =2[1/n-1/(n+1)];分别令n=1,2,3, ..., n,得n个式子;将这n个式子两边相加即得,前n项和 Sn=[1/1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+n)]-1 =2×[1-1/(n+1)]-1 =(n-1)/(n+1)。