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[函数] 证明 $f(2)\geqslant f(n)\geqslant f(3)$

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hbghlyj Posted 2025-3-1 19:25 |Read mode
Last edited by hbghlyj 2025-3-3 07:44$n\geqslant2,$
$$f(n)=n^{1/2}\prod_{k=2}^n\left(1 +\frac{(-1)^k}{\sqrt{k}}\right)$$
如何证明 $f(2)\geqslant f(n)\geqslant f(3)$ 呢?

$f(2)=1+\sqrt2$
$f(3)=-1+\sqrt{3}+\sqrt{2-\sqrt{3}}$

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 Author| hbghlyj Posted 2025-3-1 19:29

如何证明 $f(2)\geqslant f(n)\geqslant f(3)$ 呢?

  1. from math import sqrt, prod
  2. import matplotlib.pyplot as plt
  3. def f(n):
  4.   return n**(1/2) * prod(1 + (-1)**i * 1/sqrt(i) for i in range(2, n+1))
  5. x = list(range(2, 100))
  6. y = [f(n) for n in x]
  7. plt.plot(x, y)
  8. plt.xlabel('n')
  9. plt.ylabel('f(n)')
  10. plt.title('Line Plot of f(n)')
  11. plt.show()
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61d655ce-6ad1-4158-a079-b4a1c5e76009[1].png

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ZCos666 Posted 2025-3-1 21:23
分奇偶数列看就行了,即证
\[ \sqrt{\dfrac{n+2}{n}}\left(1+\dfrac{(-1)^{n+1}}{\sqrt{n+1}}\right)\left(1+\dfrac{(-1)^{n+2}}{\sqrt{n+2}}\right)\gtrless1 \]
即证$(\sqrt{n+1}\pm1)(\sqrt{n+2}\mp1)\gtrless\sqrt{n(n+1)}$
两种情况其实可以化归成$\dfrac{\sqrt{n+2}-1}{\sqrt{n+1}-1}>\sqrt{\dfrac{n+1}{n}}$
整理一下式子,发现可以“同构”
\[ n+1-\sqrt{(n+1)^2-1}<\sqrt{n+1}-\sqrt{n} \]
保证单调性,即证
\[ \sqrt{n+1}+\sqrt{n}<n+1+\sqrt{(n+1)^2-1} \]
那么往证$n<(n+1)^2-1$,显然有$n^2+n>0$成立
得证

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 Author| hbghlyj Posted 2025-3-1 21:46
ZCos666 发表于 2025-3-1 13:23
分奇偶数列看就行了,
这是否证明了$\{f(2n+1)\}$递增,$\{f(2n)\}$递减

所以 $f(2)>f(4)>f(6)>\ldots$

但如何证明 $f(2)>f(2n+1)$

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f(2)>f(2n)>f(2n+1)不就行了  Posted 2025-3-1 23:03

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 Author| hbghlyj Posted 2025-3-1 23:14
如何证明 $f(2n)>f(2n+1)$?

我懂了,$$\left(\sqrt{\frac{2n}{2n+1}} + \frac{1}{\sqrt{2n+1}}\right)^2>\frac{2n}{2n+1}+ \frac{1}{2n+1} = 1$$
$$\implies\sqrt{\frac{2n}{2n+1}} > 1 - \frac{1}{\sqrt{2n+1}}$$

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 Author| hbghlyj Posted 2025-3-2 19:21
@ZCos666 我也想知道,极限 $\lim f(n)$ 有闭形式吗?

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