Find the product of the series?

Question

Can anybody please help me to evaluate the following product?

$$\prod_{n=2}^\infty\dfrac{n^3 - 1}{n^3 + 1} $$

Answer 1

Note that $$ \frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}=\frac{(n-1)}{(n+1)}\cdot\frac{\big((n+1)^2-(n+1)+1\big)}{(n^2-n+1)}. $$ We have $$ \prod_{n=2}^N\frac{n-1}{n+1}=\frac{1\cdot 2\cdots(N-2)(N-1)}{3\cdot 4\cdots N(N+1)}=\frac{1\cdot 2}{N(N+1)}, $$ while \begin{align} \prod_{n=2}^N\frac{n^2+n+1}{n^2-n+1}&=\prod_{n=2}^N\frac{\big((n+1)^2-(n+1)+1\big)}{(n^2-n+1)}\\&=\frac{\big((N+1)^2-(N+1)+1\big)}{2^2-2+1}=\frac{N^2+N+1}{3}, \end{align} and hence $$ \prod_{n=2}^N\frac{n^3-1}{n^3+1}=\frac{1\cdot 2\cdot(N^2+N+1)}{3\cdot N(N+1)}\to \frac{2}{3}, $$ as $N\to \infty$.

Answer 2

Notice $$\frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$$ and $(n+1)^2-(n+1)+1=n^2+n+1$. Therefore we have a telescoping product

$$\prod_{n=2}^\infty \frac{n^3-1}{n^3+1}=\frac{1\cdot \color{blue}{7}}{\color{red}{3}\cdot 3}\cdot\frac{2\cdot \color{blue}{13}}{\color{red}{4}\cdot \color{blue}{7}}\cdot \frac{\color{red}{3}\cdot \color{blue}{21}}{\color{red}{5}\cdot \color{blue}{13}}\cdots=\frac{1\cdot 2}{3}=\frac{2}{3}$$

If nervous about doing cancelling like this in an infinite product (which is a valid point!), do it carefully for the partial products and end up with the same.