可以用調(diào)和級(jí)數(shù)的有限項(xiàng)的值為ln(n+1)+r（n) ,r(n)為歐拉常數(shù)
1+1/2+1/3+.1/n=ln(n+1)+r(n)
1+1/2+1/3+.1/2n=ln(2n+1)+r (2n)
兩者相減得
1/(n+1)+1/(n+2)+...1/2n=ln(2n+1)-ln(n+1)=ln[(2n+1)/(n+1)] =ln(2+(1/n))/(1+1/n)