Questions mathématiques diverses

Question 32:
Déterminer la limite suivante :
$\lim\limits_{n \to +\infty}\prod\limits_{k=2}^{n}{\dfrac{k^3-1}{k^3+1}} $

Réponse 32:
$$\begin{align*} k^3-1&=(k-1)(k^2+k+1)\\ k^3+1&=(k+1)(k^2-k+1)\\ \prod\limits_{k=2}^{n}{\dfrac{k^3-1}{k^3+1}} &=\prod\limits_{k=2}^{n}{\dfrac{k-1}{k+1}\dfrac{k^2+k+1}{k^2-k+1}}\\ &=\prod\limits_{k=2}^{n}{\dfrac{k-1}{k+1}}\prod\limits_{k=2}^{n}{\dfrac{k^2+k+1}{k^2-k+1}}\\ &=\dfrac{1.2.3...(n-1)}{3.4.5...(n+1)}\dfrac{7.13.21....(n^2+n+1)}{3.7.13...(n^2-n+1)}\\ Or\; (n-1&)^2+(n-1)+1=n^2-n+1\\ \prod\limits_{k=2}^{n}{\dfrac{k^3-1}{k^3+1}}&=\dfrac{1.2}{n(n+1)}\dfrac{n^2+n+1}{3}\\ &=\dfrac{2\left({1+\dfrac{1}{n}+\dfrac{1}{n^2}}\right)}{3\left({1+\dfrac{1}{n}}\right)} \end{align*}$$ Donc $\lim\limits_{n \to +\infty}\prod\limits_{k=2}^{n}{\dfrac{k^3-1}{k^3+1}}=\dfrac{2}{3}$