Limit of an infinite sequence

Question

Let $$P_n=\frac{2^3-1}{2^3+1}\frac{3^3-1}{3^3+1}.....\frac{n^3-1}{n^3+1}$$ Prove that $\lim_{n \to \infty}=\frac{2}{3}$. I tried factorizing them but not much canceling or some other thing happened. What can I do?

Answer 1

We have

\begin{align}P_n = \prod_{j = 2}^n \frac{j^3 - 1}{j^3 + 1} &= \prod_{j = 2}^n \frac{j-1}{j+1} \frac{j^2 + j + 1}{j^2 - j + 1}\\ &= \frac{2(n-1)!}{(n+1)!}\prod_{j = 2}^n \frac{(j+1)^2 - (j+1) + 1}{j^2 - j + 1}\\ &= \frac{2}{(n+1)n}\frac{n^2 + n + 1}{2^2 - 2 + 1}\quad \text{by telescoping}\\ &= \frac{2}{3}\frac{n^2+n+1}{n^2+n}\\ &\to \frac{2}{3} \quad \text{as} \quad n\to \infty. \end{align}