Limit of product with cubes $\lim\limits_{n \to \infty}\frac{2^3-1}{2^3+1}\frac{3^3-1}{3^3+1}\dots\frac{n^3-1}{n^3+1}$

Question

I am trying to evaluate $$\lim\limits_{n \to \infty}{2^3-1 \over 2^3+1}{3^3-1 \over 3^3+1}\dots{n^3-1 \over n^3+1}$$ It seems to be the perfect candidate for a factoring formula, however, I get stuck rewriting the fractions. Do you have any suggestions?

Answer 1

One may use a telescoping product, as $n \to \infty$, $$ \prod_{k=2}^n\frac{k^3-1}{k^3+1}=\prod_{k=2}^n\frac{k-1}{k+1} \cdot \prod_{k=2}^n\frac{k(k+1)+1}{k(k-1)+1}=\frac23 \cdot \frac{n(n+1)+1}{n(n+1)}\to \frac23. $$

Answer 2

$$ \begin{align} \prod_{k=2}^\infty\frac{k^3-1}{k^3+1} &=\lim_{n\to\infty}\prod_{k=2}^n\frac{(k-1)\left(k^2+k+1\right)}{(k+1)\left(k^2-k+1\right)}\\ &=\lim_{n\to\infty}\underbrace{\prod_{k=2}^n\frac{k-1}{k+1}}\underbrace{\prod_{k=2}^n\frac{k(k+1)+1}{(k-1)k+1}}\\ &=\lim_{n\to\infty}\,\frac{1\cdot2}{n(n+1)}\quad\frac{n(n+1)+1}{3}\\[3pt] &=\frac23 \end{align} $$