> 数学 >
等差数列{an}，{bn}的前n项和分别为Sn和Tn，若
Sn
Tn
=
2n
3n+1
，则
lim
n→∞
an
bn
=______．
人气：278 ℃　时间：2019-08-19 03:32:29
解答
∵
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
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版