Infinite product: $\prod_{k=2}^{n}\frac{k^3-1}{k^3+1}$

This question already has an answer here:

  • How to compute $\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$

    3 answers

Solutions Collecting From Web of "Infinite product: $\prod_{k=2}^{n}\frac{k^3-1}{k^3+1}$"

The main idea is to use a telescope product of the form $ \prod \frac{b_j}{b_{j+2}} \frac{a_{j+1}}{a_j}$. We have$$ \lim_{n \to \infty} \prod_{k=2}^n \frac{k^3-1}{k^3+1}=\lim_{n \to \infty}\prod_{k=2}^n \frac{k-1}{k+1} \frac{k^2+k+1}{k^2-k+1}=\lim_{n \to \infty}\prod_{k=2}^n \frac{k-1}{(k+2)-1} \frac{(k+1)^2-(k+1)+1}{k^2-k+1}=\lim_{n \to \infty} \frac{1 \cdot 2}{n \cdot (n+1)}\frac{n^2+n+1}{4-2+1}=\lim_{n \to \infty} \frac{2}{3} \frac{n^2+n+1}{n^2+n}=\frac{2}{3}.$$