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

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

My Working :

$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}= 1 – \prod^{\infty}_{n=2}\frac{2}{n^3+1}
= 1-0 = 1$$

Is it correct

Solutions Collecting From Web of "How to compute $\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$"

Magic answer:

Let $f(n) =\dfrac{n(n-1)}{n^2-n+1}$. Then show $f(n+1) = \dfrac{n(n+1)}{n^2+n+1}$ and thus $$\frac{f(n)}{f(n+1)} = \frac{n(n-1)(n^2+n+1)}{n(n+1)(n^2-n+1)} = \frac{n^3-1}{n^3+1}$$

(I call this a “magic answer” just because most of the other answers here give you reasons for how you would see this, while I just plop in an $f$ that works, as if by magic. This is really the same argument as the others, just distilled to a minimalist essence.)


If $$t_n=\frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$$


and $$t_{n-1}=\frac{(n-2)\{(n-1)^2+n-1+1\}}{n\{(n-1)^2-(n-1)-1\}}=\frac{(n-2)(n^2-n+1)}{n\{(n-1)^2-(n-1)-1\}}$$

Alternatively, let $\displaystyle u_n=\frac{n-1}{n+1},v_n=\frac{n^2+n+1}{n^2-n+1}$ so that $t_n=u_n\cdot v_n$

$$\implies\prod_{2\le n\le r}u_n=\frac{1\cdot2\cdot3\cdots(r-2)(r-1)}{3\cdot4\cdot5\cdots r(r+1)}\frac2{r(r+1)}$$

$$\implies\prod_{2\le n\le r}v_n=\frac{7\cdot13\cdot21\cdots(r^2-r+1)(r^2+r+1)}{3\cdot7\cdot13\cdots \{(r-1)^2-(r-1)+1\}(r^2-r+1)}=\frac{r^2+r+1}3$$

$$\implies\prod_{2\le n\le r} t_n=\left(\prod_{2\le n\le r}u_n\right)\left( \prod_{2\le n\le r}v_n\right)=\frac{2(r^2+r+1)}{3r(r+1)} $$

Setting $r\to\infty,$ $\displaystyle \frac{2(r^2+r+1)}{3r(r+1)}=\frac23\lim_{r\to\infty}\frac{1+\frac1r+\frac1{r^2}}{1+\frac1r}=?$

Note that
$${n^3 – 1 \over n^3 + 1} = {n – 1 \over n + 1}{n^2 + n + 1 \over n^2 – n + 1}$$
Also note that
$$(n-1)^2 + (n – 1) + 1 = n^2 – n + 1$$
So the infinite product in question is really the product of two telescoping products.