Infinite Products!

After simplifying a little, we get the following: $\frac{f(6)}{f^2(3)} = \frac{\prod_{n=2}^{\infty} \frac{n^6 - 1}{n^6}}{\prod_{n=2}^{\infty} \frac{(n^3 - 1)^2}{n^6}} =\prod_{n=2}^{\infty} \frac{n^6 - 1}{(n^3-1)^2} =\prod_{n=2}^{\infty} \frac{n^3 + 1}{n^3-1}$ $= \prod_{n=2}^{\infty} \frac{(n+1)(n^2-n+1)}{(n-1)(n^2+n+1)}$ Now let's use partial products to help us figure this out. Let $P_m$ represent the $m^{th}$ partial product for $m \ge 2$ : $P_m = \prod_{n=2}^{m} \frac{(n+1)(n^2-n+1)}{(n-1)(n^2+n+1)} = \prod_{n=2}^{m} \frac{n+1}{n-1} \prod_{n=2}^{m} \frac{(n-1)^2+(n-1)+1}{n^2+n+1}$ So now we have two telescoping products. It can be shown by induction that the first product is equal to $\dfrac{m(m+1)}{2}$ , and the second is equal to $\dfrac{3}{m^2+m+1}$ . Therefore, $P_m = \dfrac{m(m+1)}{2}\dfrac{3}{m^2+m+1} = \frac{3}{2}\left(1 - \dfrac{1}{m^2+m+1}\right)$ Therefore, as $m$ approaches infinity, this product approaches $\boxed{\dfrac{3}{2}}$