[M@gpie]

เนื่องจาก \[ \frac{n^3-1}{n^3+1} = \frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} \]
\[ \prod_{n=2}^{k}\frac{n^3-1}{n^3+1} = \prod_{n=2}^{k}\frac{(n-1)}{(n+1)} \prod_{n=2}^{k}\frac{n^2+n+1}{n^2-n+1} = \frac{2}{3} \frac{(k+1)^2-(k+1)+1}{k(k+1)}\]
\[ \]
\[ lim_{k \rightarrow \infty}\prod_{n=2}^{k}\frac{n^3-1}{n^3+1} = \frac{2}{3} \]