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[数列] 数列的单调性

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睡神 Post time 2024-5-2 10:20 From the mobile phone |Read mode
证明或否定:数列$a_n=\left(\dfrac{1}{n}\right)^n+\left(\dfrac{2}{n}\right)^n+\cdots+\left(\dfrac {n}{n}\right)^n$单调递增
除了不懂,就是装懂

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realnumber Post time 2024-5-3 06:47
只要证,对任意给定$i\in N$,$i\le n$有$(\frac{i}{n})^n\le (\frac{i+1}{n+1})^{n+1}$.两边对i求和,即证明了单调递增.
记$f(n)=(n+1)\ln \frac{i+1}{n+1}-n\ln \frac{i}{n}$,求一下导
即要证$f'(n)=\ln \frac{i+1}{n+1}-\ln \frac{i}{n}\ge 0$成立.

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 Author| 睡神 Post time 2024-5-3 08:09
realnumber 发表于 2024-5-3 06:47
只要证,对任意给定$i\in N$,$i\le n$有$(\frac{i}{n})^n\le (\frac{i+1}{n+1})^{n+1}$.两边对i求和,即证明 ...

$(\dfrac{i}{n})^n\le (\dfrac{i+1}{n+1})^{n+1}$这个好像反向了吧?莫非我算错了?

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至少i=0,1没反向  Post time 2024-5-3 08:38
检查了下,好像没错,要不你也写几步?  Post time 2024-5-3 09:36
除了不懂,就是装懂

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 Author| 睡神 Post time 2024-5-3 10:11 From the mobile phone
realnumber 发表于 2024-5-3 06:47
只要证,对任意给定$i\in N$,$i\le n$有$(\frac{i}{n})^n\le (\frac{i+1}{n+1})^{n+1}$.两边对i求和,即证明 ...

噢,重新算了一下,这个是对的!之前我也考虑过这个,可能当时我算错了,非常抱歉😂感谢大佬!
除了不懂,就是装懂

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