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Question:

How prove that :$$\prod_{k=2}^n \left({\frac{k^3-1}{k^3+1}} \right) \le\frac{3}{2}$$

My seggestion: we know that : $(k^3+1)=(k+1)(k^2-k+1)$ $(k^3-1)=(k-1)(k^2+k+1)$

This implies that : $$ \prod_{k=2}^n \left({\frac{k^3-1}{k^3+1}} \right) = \prod_{k=2}^n \left({\frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)}} \right) = \prod_{k=2}^n \left({\frac{k-1}{k+1}}\right)\prod_{k=2}^n \left({\frac{k^2+k+1}{k^2-k+1}} \right) $$

and we know that : $$\prod_{k=2}^n \left({\frac{k-1}{k+1}}\right)=\frac {3}{1}\cdot\frac {4}{2}\cdot\frac {5}{3}\cdot\frac {6}{4}\cdot\cdot\cdot\frac {n-2}{n-4}\cdot\frac {n-1}{n-3}\cdot\frac {n}{n-2}\cdot\frac {n＋1}{n-1}＝\frac {n(n＋1)}{2}$$ and therfore : $$\prod_{k=2}^n \left({\frac{k^3-1}{k^3+1}} \right) =\frac{n(n+1)}{2}\prod_{k=2}^n \left({\frac{k^2+k+1}{k^2-k+1}} \right) $$

Mostafa
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    What about $\frac{k^3-1}{k^3+1} <\frac{k^3-1}{k^3+1} + \frac{2}{k^3+1} = 1$? Doesn't this give a straightforward solution? – Yuumita Jul 27 '23 at 12:21
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    Notice that the second product is telescopic : $$\prod_{k = 2}^n \dfrac{k^2 + k + 1}{k^2 - k + 1} = \prod_{k = 2}^n \dfrac{(k + 1)^2 - (k + 1) + 1}{k^2 - k + 1} = \dfrac{(n + 1)^2 - (n + 1) + 1}{2^2 - 2 + 1} = \dfrac{n^2 + n + 1}{3}$$ – Essaidi Jul 27 '23 at 12:32
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    $$\prod_{k = 2}^n \dfrac{k - 1}{k + 1} = \dfrac{\prod_{k = 2}^n (k - 1)}{\prod_{k = 2}^n (k + 1)} = \dfrac{\prod_{k = 1}^{n - 1} k}{\prod_{k = 3}^{n + 1} k} = \dfrac{2}{n(n + 1)}$$ – Essaidi Jul 27 '23 at 12:39
  • May be you want to show that the product is $\leq \dfrac{2}{3}$ and not $\dfrac{3}{2}$. – Essaidi Jul 27 '23 at 12:44

1 Answers1

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As Yuumita suggested, $\frac{k^3−1}{k^3+1}<\frac{k^3−1}{k^3+1}+\frac{2}{k^3+1}=1$. Furthermore, this product tends to $\frac{2}{3}$. So if that was what you're actually looking for, there are a few answers that can help you with that.

Sgg8
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