Prove that $\prod\limits_{n= 2}^{\infty} \frac{n^{3}+ 1}{n^{3}- 1}= \frac{3}{2}$, which is a beautiful value!

Question

Prove that $\prod\limits_{n= 2}^{\infty} \dfrac{n^{3}+ 1}{n^{3}- 1}= \dfrac{3}{2}$, which is a beautiful value ! A solution due to a p u as follow $$\begin{align} \prod\limits_{n= 2}^{\infty} \frac{n^{3}+ 1}{n^{3}- 1} & = \prod\limits_{n= 2}^{\infty}\frac{(n+ 1)(n+ \omega )(n+ \omega^{2})}{(n- 1)(n- \omega )(n- \omega^{2})}\\ & = \prod\limits_{n= 2}^{\infty} \frac{(n+ 1)(n- 1- \omega)(n- 1- \omega^{2})}{(n- 1)(n- \omega )(n- \omega^{2})}\,(\!1+ \omega+ \omega^{2}= 0\,used\!)\\ & = \lim_{n\rightarrow \infty} \frac{n(n+ 1)(1- \omega)(1- \omega^{2})}{1\cdot 2(n- \omega)(n- \omega^{2})}\\ & = \lim_{n\rightarrow \infty} \frac{3(1+ \frac{1}{n})}{2(1- \frac{\omega}{n})(1- \frac{\omega^{2}}{n})}\,\left (\!(1- \omega)(1- \omega^{2})= 1+ 1+ 1= 3\,used\!\right )\\ & = \frac{3}{2} \end{align}$$ I have a solution, and I'm looking forward to seeing a nicer one(s), thanks for your interests a lot !

Answer 1

Here is a much simpler way. Observe that the partial product

$$ \prod_{n=2}^{m} \frac{n^3+1}{n^3-1} = \frac{3 (m^2+m)}{2(m^2+m+1)}. $$

Then when $m\to\infty$, the product converges to $\displaystyle \frac{3}{2}$.