∵1/(1+2+3+...+k)=2/[k(k+1)] (k=1,2,3,.) (应用等差数列求和公式)=2[1/k-1/(k+1)]∴1/(1+2)=2(1/2-1/3)1/(1+2+3)=2(1/3-1/4).1/(1+2+3+.+n)=2[1/n-1/(n+1)]故 lim(n->∞)[(1/(1+2)+1/(1+2+3)+.+1/(1+2+3+...+n)=li...