Infinite product 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 

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 $\frac{n^2+n+1}{n^2-n+1}$ .
