How do I solve the infinite product of $$\prod_{n=2}^\infty\frac{n^3-1}{n^3+1}?$$
I know that I have to factorise to $$\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)},$$
but how do I do the partial product?
Thanks a lot in advance.
If I'm not mistaken the Answer is 2/3
On
Hint: Let $f(x)=x^2+x+1$. Then $f(n)=n^2+n+1$, and $f(n-1)=n^2-n+1$. This will enable you to "telescope" the terms $\frac{n^2+n+1}{n^2-n+1}$. A whole lot of cancelling going on.
The $\frac{n-1}{n+1}$ terms also telescope.
I would suggest that you write down the terms you are multiplying, for $n=2$, $3$, $4$, even $5$. Express each term in the factored form mentioned in the post. For example, for $n=2$ we will have $\frac{1\cdot 7}{3\cdot 3}$. The collapse will be visually clear.
After factorization, the product looks like $(\frac{1}{3}\frac{2}{4}\frac{3}{5}\frac{4}{6}\frac{5}{7}\cdots)(\frac{7}{3}\frac{13}{7}\frac{21}{13}\frac{31}{21}\frac{43}{31}\cdots)=(2)(1/3)=2/3$.$$$$ Here terms in first () are from expression $\frac{n-1}{n+1}$ and terms in second () from expression $\frac{n^2+n+1}{n^2-n+1}$ .