看看通项(1/n)/[(1+1/2)(1+1/3)…(1+1/n)]=(1/n)/[(3/2)*(4/3)*…*(n+1)/n]=(1/n)/[(n+1)/2]=2/[n(n+1)]=2[1/n-1/(n+1)]故，原式=2(1/2-1/3)+2(1/3-1/4)+…+2(1/2005-1/2006)=2(1/2-1/2006)=1-1/1003=1002/1003
请帮一个有上进心但成绩较差一点的同学解答
