题目内容
设数列{an}的各项均为正数,前n项和为Sn,对于任意的n∈N+,an,Sn,a成等差数列,设数列{bn}的前n项和为Tn,且bn=
,若对任意的实数x∈(1,e](e是自然对数的底)和任意正整数n,总有Tn<r(r∈N+).则r的最小值为________.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824034612892701.png)
2
根据题意,对于任意n∈N+,总有an,Sn,a成等差数列,则对于n∈N*,总有2Sn=an+
①
所以2Sn-1=an-1+
(n≥2)②
①-②得2an=an+
-an-1-
,即an+an-1=(an+an-1)(an-an-1)因为an,an-1均为正数,所以an-an-1=1(n≥2),
所以数列{an}是公差为1的等差数列,又n=1时,2S1=a1+a,解得a1=1,所以an=n,对于任意的实数x∈(1,e],有0<ln x<1,对于任意正整数n.总有bn=
≤
,所以Tn≤
=
,又对任意的实数x∈(1,e]和任意正整数n,总有Tn<r(r∈N+),所以r的最小值为2.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824034612923398.png)
所以2Sn-1=an-1+
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824034612939445.png)
①-②得2an=an+
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824034612923398.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824034612939445.png)
所以数列{an}是公差为1的等差数列,又n=1时,2S1=a1+a,解得a1=1,所以an=n,对于任意的实数x∈(1,e],有0<ln x<1,对于任意正整数n.总有bn=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824034612892701.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824034613001414.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240346130171595.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240346130321195.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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