<s>[INST] 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?
 [/INST] One may use a telescoping product, as $n \to \infty$,
$$
\prod_{k=2}^n\frac{k^3-1}{k^3+1}=\prod_{k=2}^n\frac{k-1}{k+1} \cdot \prod_{k=2}^n\frac{k(k+1)+1}{k(k-1)+1}=\frac23 \cdot \frac{n(n+1)+1}{n(n+1)}\to \frac23.
$$
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