题目内容
如果正数数列{an}满足:对任意的正数M,都存在正整数n,使得![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_ST/0.png)
(Ⅰ)若an=3+2sin(n)(n=1,2,3,…),
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_ST/1.png)
(Ⅱ)若an=n+2,是否存在正整数k,使得对于一切n≥k,有
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_ST/2.png)
(Ⅲ)若数列{an}是单调递增的无界正数列,求证:存在正整数m,使得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_ST/3.png)
【答案】分析:(Ⅰ)取M=5,显然an=3+2sin(n)≤5不符合无界正数列的定义;对任意的正数M,取n为大于2M的一个偶数,
符合无界正数列的定义.
(Ⅱ)
变形为
从而求得;
(Ⅲ)观察要证的不等式的结构与(II)相似,故应用(II)变形后,再由{an}是单调递增的无界正数列证明.
解答:解:(Ⅰ){an}不是无界正数列.理由如下:
取M=5,显然an=3+2sin(n)≤5,不存在正整数n满足
;{bn}是无界正数列.理由如下:
对任意的正数M,取n为大于2M的一个偶数,有
,所以{bn}是无界正数列.
(Ⅱ)存在满足题意的正整数k.理由如下:
当n≥3时,
因为
=
=
,
即取k=3,对于一切n≥k,有
成立.
注:k为大于或等于3的整数即可.
(Ⅲ)证明:因为数列{an}是单调递增的正数列,
所以
=![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/10.png)
.
即
.
因为{an}是无界正数列,取M=2a1,由定义知存在正整数n1,使
.
所以
.
由定义可知{an}是无穷数列,考察数列
,
,
,
显然这仍是一个单调递增的无界正数列,同上理由可知存在正整数n2,使得
.
重复上述操作,直到确定相应的正整数n4018.
则
=n4018-2009.
即存在正整数m=n4018,使得
成立.
点评:本题通过情境设置定义新的数列在研究中渗透着不等式的构造、变形、放缩,培养学生灵活运用知识的能力.
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/0.png)
(Ⅱ)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/2.png)
(Ⅲ)观察要证的不等式的结构与(II)相似,故应用(II)变形后,再由{an}是单调递增的无界正数列证明.
解答:解:(Ⅰ){an}不是无界正数列.理由如下:
取M=5,显然an=3+2sin(n)≤5,不存在正整数n满足
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/3.png)
对任意的正数M,取n为大于2M的一个偶数,有
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/4.png)
(Ⅱ)存在满足题意的正整数k.理由如下:
当n≥3时,
因为
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/6.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/7.png)
即取k=3,对于一切n≥k,有
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/8.png)
注:k为大于或等于3的整数即可.
(Ⅲ)证明:因为数列{an}是单调递增的正数列,
所以
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/11.png)
即
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/12.png)
因为{an}是无界正数列,取M=2a1,由定义知存在正整数n1,使
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/13.png)
所以
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/14.png)
由定义可知{an}是无穷数列,考察数列
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/15.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/16.png)
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/17.png)
显然这仍是一个单调递增的无界正数列,同上理由可知存在正整数n2,使得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/18.png)
重复上述操作,直到确定相应的正整数n4018.
则
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/19.png)
即存在正整数m=n4018,使得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131101221415432372829/SYS201311012214154323728019_DA/20.png)
点评:本题通过情境设置定义新的数列在研究中渗透着不等式的构造、变形、放缩,培养学生灵活运用知识的能力.
![](http://thumb2018.1010pic.com/images/loading.gif)
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