Let $P_n=\frac{2^3-1}{2^3+1} \cdot \frac{3^3-1}{3^3+1} \cdot\frac{4^3-1}{4^3+1}...\frac{n^3-1}{n^3+1}$
Then $\lim_{n\to\infty} P_n=\frac{2}{3}$.
I could understand that as $n \to \infty$, the last term in the series approaches $1$. But I couldn't understand how the product evaluates to $\frac{2}{3}$.
To give a gist of the answers: We have $$ \lim_{n \to \infty} \prod_{a=2}^n \frac{a^3-1}{a^3+1}=\lim_{n \to \infty}\prod_{a=2}^n \frac{a-1}{a+1} \frac{a^2+a+1}{a^2-a+1}=\lim_{n \to \infty}\prod_{a=2}^n \frac{a-1}{(a+2)-1} \frac{(a+1)^2-(a+1)+1}{a^2-a+1}=\lim_{n \to \infty} \frac{1 \cdot 2}{n \cdot (n+1)}\frac{n^2+n+1}{4-2+1}=\lim_{n \to \infty} \frac{2}{3} \frac{n^2+n+1}{n^2+n}=\frac{2}{3}$$ Hope it helps.