Show that $\prod_{k=2}^{\infty} \left(\frac{k^3-1}{k^3+1}\right)=\frac{2}{3}$
Ok so in Remmert's Classical Topics in Complex Function Theory In the introduction to infinite products, there is this excercise.
And Im having trouble in trying to compute the partial products. Im not sure if there is a standard way in trying to solve this problems or if it is just simple intuition and trying. Is there any rule? or method as to solve this kind of products?
Since $$k^3-1=(k-1)(k^2+k+1),\quad k^3+1=(k+1)(k^2-k+1)$$ we have $$\begin{align}\lim_{n\to\infty}\prod_{k=2}^{n}\frac{k^3-1}{k^3+1}&=\lim_{n\to \infty}\prod_{k=2}^{n}\frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)}\\&=\lim_{n\to\infty}\prod_{k=2}^{n}\frac{k-1}{k+1}\prod_{k=2}^{n}\frac{(k+1)^2-k}{k^2-(k-1)}\\&=\lim_{n\to\infty}\frac{2}{n(n+1)}\cdot\frac{(n+1)^2-n}{2^2-(2-1)}\\&=\lim_{n\to\infty}\frac{2+\frac 2n+\frac{2}{n^2}}{3+\frac 3n}\\&=\frac 23\end{align}$$