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[不等式] $\frac{1}{1^2}+\frac{1}{3^2}+\cdots+\frac{1}{(2n-1)^2} < \frac{5}{4}$

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郝酒 Posted 2015-6-14 20:41 |Read mode
想问下这个除了基本不等式放缩,还有没有其他的证法?

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 Author| 郝酒 Posted 2015-6-16 16:09
顶顶,看有没有别的证法。

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活着&存在 Posted 2015-6-16 20:36
未命名.JPG

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战巡 Posted 2015-6-16 21:56
回复 1# 郝酒


\[\sum_{k=1}^{n}\frac{1}{(2k-1)^2}<\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}=\zeta(2)-\frac{1}{2^2}\zeta(2)=\frac{\pi^2}{6}·\frac{3}{4}=\frac{\pi^2}{8}\]

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 Author| 郝酒 Posted 2015-6-18 13:08
求证 $\zeta(2)=\frac{\pi^2}{6}$

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战巡 Posted 2015-6-18 14:14
回复 5# 郝酒


这是常识,自己看书

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 Author| 郝酒 Posted 2015-6-18 20:58
Last edited by 郝酒 2015-6-19 23:46这跟现代人用智能手机是一个类型……
$\zeta(2)=\pi^2/6$并不是一个比放缩更简单的问题。

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战巡 Posted 2015-6-19 11:40
回复 7# 郝酒


下次劳资就把题目改成让你证明这个玩意$<\frac{\pi^2}{8}$,看你还怎么放缩

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 Author| 郝酒 Posted 2015-6-19 23:47
只要严格小于就有放缩的空间。
话说这种放缩的技巧还是有教育的价值的。

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