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[不等式] 证$\frac 1{n+1}+\frac 1{n+2}+\cdots+\frac 1{3n+1}<\frac 98$

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isee Posted 2014-9-19 15:20 |Read mode
设$n$为自然数,求证\[\frac 1{n+1}+\frac 1{n+2}+\cdots+\frac 1{3n+1}<\frac 98.\]

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爪机专用 Posted 2014-9-19 15:55
又是这种,已经玩烂了。
直接求极限吧

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爪机专用 Posted 2014-9-19 16:14
可以参考这个:forum.php?mod=viewthread&tid=1736

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 Author| isee Posted 2014-9-19 20:41
回复 2# 爪机专用


    说明这种题的确是难度大

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战巡 Posted 2014-9-20 03:29
回复 4# isee


出烂了的题难度就不会大,早就有套路完爆它了
一眼看出极限为$ln(3)<\frac{9}{8}$

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 Author| isee Posted 2014-9-20 09:55
回复 5# 战巡


    有没有具体的过程,我学习一下,,,谢谢

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tommywong Posted 2014-9-20 10:28
回复 6# isee


    $\displaystyle \int_1^{2n+1} \frac{1}{n+x} dx$

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kuing Posted 2014-9-20 11:47
回复 6# isee

根本没看3#链接……

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其妙 Posted 2014-9-20 15:13
不使用牛莱公式和欧拉常数来证明

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其妙 Posted 2014-9-21 14:42
简单的不等式:设$n$为自然数,$n\geqslant2$,求证$\frac 1{n}+\frac 1{n+1}+\frac 1{n+2}+\cdots+\frac 1{n^2}>1$

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 Author| isee Posted 2022-1-14 17:55
回复  isee

根本没看3#链接……
kuing 发表于 2014-9-20 11:47
哈哈哈,当年吧,直接给忽略了~

在 $\ln (x+1)<x$ 中分别令 $x=\frac 1{n}, x=-\frac 1{n}$ 有
$$\ln \frac {n+1}{n}<\frac 1{n}<\ln \frac {n}{n-1}.$$
于是
\begin{align*} \ln \frac {n+2}{n+1}&<\frac 1{n+1}<\ln \frac {n+1}{n}\\[1em]
\ln \frac {n+3}{n+2}&<\frac 1{n+2}<\ln \frac {n+2}{n+1}\\[1em]
&\cdots\qquad \qquad \cdots\\[1em]
\ln \frac {3n+2}{3n+1}&<\frac 1{n+2n+1}<\ln \frac {3n+1}{3n}
\end{align*}
这 $2n+1$ 个式子相加,便有
\begin{align*} \ln \frac {3n+2}{3n+1}&<\frac 1{n+1}+\frac 1{n+2}+\cdots+\frac 1{3n+1}<\ln \frac {3n+1}{n}, \end{align*}
由夹逼法则,知
$$\lim_{n \to \infty}\frac 1{n+1}+\frac 1{n+2}+\cdots+\frac 1{3n+1}=\ln 3.$$

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 Author| isee Posted 2022-1-14 19:09
回复  isee


    $\displaystyle \int_1^{2n+1} \frac{1}{n+x} dx$
tommywong 发表于 2014-9-20 10:28
是指 和式 小于这个 积分式 吧?

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 Author| isee Posted 2022-1-14 20:07
回复 5# 战巡

现在回头一看,结果就在5#


又一个源自知乎提问,好像是个高考题采用过的

: $\lim_{n \to \infty}\frac 1{4n+1}+\frac 1{4n+2}+\cdots+\frac 1{4n+n}.$

还可以利用 $\ln (x+1)<x$ 来写,高中生都可以看明白.

在 $\ln (x+1)<x$ 中分别令 $$x=\frac 1{4n},x=-\frac 1{4n}$$ 有

$$\ln \frac {4n+1}{4n}<\frac 1{4n}<\ln \frac {4n}{4n-1}.$$

于是

\begin{align*} \ln \frac {4n+2}{4n+1}&<\frac 1{4n+1}<\ln \frac {4n+1}{4n}\\[1em] \ln \frac {4n+3}{4n+2}&<\frac 1{4n+2}<\ln \frac {4n+2}{4n+1}\\[1em] &\cdots\qquad \qquad \cdots\\[1em] \ln \frac {5n+1}{4n+n}&<\frac 1{4n+n}<\ln \frac {5n}{5n-1} \end{align*}

这 $n$ 个式子相加,便有

\begin{align*} \ln \frac {5n+1}{4n+1}&<\frac 1{4n+1}+\frac 1{4n+2}+\cdots+\frac 1{4n+n}<\ln \frac 54, \end{align*}

由夹逼法则,知

$$\lim_{n \to \infty}\frac 1{4n+1}+\frac 1{4n+2}+\cdots+\frac 1{4n+n}=\ln \frac 54.$$

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 Author| isee Posted 2022-1-14 20:08
主楼怎么转化为定积分呢,疑问中

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