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

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