Progression

$\displaystyle \lim_{n\to \infty} 3a_n$

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

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

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

$\displaystyle= 3\frac{\prod_{n=0}^{\infty} (n+1) }{\prod_{n=2}^{\infty} (n+1) } \frac{\prod_{n=3}^{\infty} ((n-1)(n)+1)}{\prod_{n=2}^{\infty} ((n-1)(n)+1) }$

$\displaystyle = 3(0+1)(1+1) \frac{1}{(2-1)(2) +1 } = \boxed{2}$

let's consider the last three terms

$\frac{(n-1)^{3}-1}{(n-1)^{3}+1} \cdot\frac{(n)^{3}-1}{(n)^{3}+1}\cdot\frac{(n+1)^{3}-1}{(n+1)^{3}+1}$

$\frac{(n-2)(n^{2}-n+1)}{n(n^{2}-3n+3)}\cdot\frac{(n-1)(n^{2}+n+1)}{(n+1)(n^{2}+n+1)}\cdot\frac{n(n^{2}+3n+3)}{(n+2)(n^{2}+n+1)}$

so you can see that we can eliminate some terms and if we keep continuing we'll get

$\frac{2}{3}\cdot\frac{n^{2}+3n+3}{(n+1)(n+2)}$

and in the limit that $\lim_{n-> \infty} a_{n}=\frac{2}{3}$

then $\lim_{n-> \infty} 3a_{n}=2$