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[数列] 數列不等式a(m+1)=a(m)/m+m/a(m)

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tommywong posted 2018-4-2 15:07 |Read mode
Last edited by tommywong 2018-4-4 07:42这是台湾网友 YAG 发表在“陆老师的《数学中国》园地”的一个帖子,

欢迎大家一起来想想如何解答:


mathchina.net/dvbbs/dispbbs.asp?boardid=2&Id=11841

$\displaystyle a_1=1,a_{m+1}=\frac{a_m}{m}+\frac{m}{a_m},\text{證明}\sqrt{2017}<a_{2017}<\frac{2017}{\sqrt{2016}}$
现充已死,エロ当立。
维基用户页:https://zh.wikipedia.org/wiki/User:Tttfffkkk
Notable algebra methods:https://artofproblemsolving.com/community/c728438
《方幂和及其推广和式》 数学学习与研究2016.

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zhcosin posted 2018-4-3 11:27
回复 1# tommywong
没难度,用数学归纳法证明下式即可:
\[ \sqrt{n} <a_{n} < \sqrt{n-1}+\frac{1}{\sqrt{n-1}} \]

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original poster tommywong posted 2018-4-3 20:10
回复 2# zhcosin

$\displaystyle \sqrt{m}<a_m<\frac{m}{\sqrt{m-1}}?$

$\displaystyle \sqrt{m+1}>\frac{1}{\sqrt{m}}+\sqrt{m-1}<\frac{a_m}{m}+\frac{m}{a_m}<\frac{1}{\sqrt{m-1}}+\sqrt{m}>\frac{m+1}{\sqrt{m}}$

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zhcosin posted 2018-4-3 22:47
Last edited by zhcosin 2018-4-3 23:15 Selection_013.png

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original poster tommywong posted 2018-4-4 07:20
谢谢楼上 zhcosin 的解答。我已将帖子转贴到“数学中国论坛”,然后再让陆老师转贴到“陆老师的《数学中国》园地”。

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zhcosin posted 2018-4-4 08:08
回复 5# tommywong

一个题目而已,搞这么大阵仗干啥,又不是xx大定理。

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其妙 posted 2018-4-5 22:11
回复  tommywong

一个题目而已,搞这么大阵仗干啥,又不是xx大定理。
zhcosin 发表于 2018-4-4 08:08
这次和扎克风度相似

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