(an+bn)=n(n+1)+(n+1)^2
=(n+1)(2n+1)>2n(n+1)
1/(an+bn)=1/(n+1)(2n+1)
=2/(2n+1)-2/(2n+2)
从第二项开始放缩,即1/(an+bn)=1/(n+1)(2n+1)< 1/2n(n+1)=1/2（1/n-1/n+1）
1/(a1+b1)+1/(a2+b2)+1/(a3+b3)+……+1/(an+bn)<
2/3-2/4+1/2(1/2-1/3+1/3-1/4+.+1/n-1/n+1
=1/6+1/2(1/2-1/n+1)