等差数列{an}，{bn}的前n项和分别为Sn和Tn，若SnTn=2n/3n+1，则limn→∞anbn=_．
问题描述：
等差数列{an}，{bn}的前n项和分别为Sn和Tn，若
=Sn Tn
，则2n 3n+1
lim n→∞
=______． an bn
答
∵
=Sn Tn
，2n 3n+1
∴
=an bn
=a1+a2n−1 b1+b2n−1
=S2n−1 T2n−1
=2(2n−1) 3(2n−1)+1
2n−1 3n−1
∴
lim n→∞
=an bn
lim n→∞
=2n−1 3n−1
lim n→∞
=2−
1 n 3−
1 n
2 3
故答案为：
2 3