Problem: What is $\prod_{k = 2}^{\infty} \frac{k^3 - 1}{k^3 + 1}$?
My Approach: We can rewrite as $$\prod_{k = 2}^{\infty} \frac{k^3 - 1}{k^3 + 1} = \left(\prod_{k = 2}^{\infty} \frac{k - 1}{k + 1}\right)\left(\prod_{k = 2}^{\infty} \frac{k^2 + k + 1}{k^2 - k + 1}\right).$$ We can see the first product approaches by pattern as $$ \prod_{k = 2}^{\infty} \frac{k - 1}{k + 1} = \left(\frac{1}{3}\right)\left(\frac{2}{4}\right)\left(\frac{3}{5}\right)\left(\frac{4}{6}\right)\cdots = \lim_{N \to \infty}\frac{2}{N}.$$ Similarly, $$\prod_{k = 2}^{\infty} \frac{k^2 + k + 1}{k^2 - k + 1} = \left(\frac{7}{3}\right)\left(\frac{13}{7}\right)\left(\frac{21}{13}\right)\cdots = \lim_{N\to\infty}\frac{N}{3}.$$ Therefore, $$\prod_{k = 2}^{\infty} \frac{k^3 - 1}{k^3 + 1} = \lim_{N \to \infty}\frac{2}{N}\times \lim_{N\to\infty}\frac{N}{3} = \lim_{N\to\infty}\frac{2}{3} = \frac{2}{3}.$$ This doesn't feel rigorous enough to me because I'm just looking at the pattern. Is there anything I can do to make this a better proof?