Find $\lim_{n \rightarrow\infty } P_n$whreas $P_n=\frac{2^3-1}{2^3+1}\cdot\frac{3^3-1}{3^3+1}\cdot\cdot\cdot\frac{n^3-1}{n^3+1}$. $\lim_{n \rightarrow\infty }  P_n$whreas $P_n=\frac{2^3-1}{2^3+1}\cdot\frac{3^3-1}{3^3+1}\cdot\cdot\cdot\frac{n^3-1}{n^3+1}$.
This is a past problem of a high-school level math compettion.
My tries:
1.Initially I thought of finding the product of the sequence.But then I realize there is no need of product coz' it'll make the function more complicated.
2.Then I tried to simplify first few terms so that may be in some way some of the middle terms may be cancelled out .
So the product of first few terms look sth like this,
$\frac{7}{9}\cdot\frac{26}{28}\cdot\frac{63}{65}\cdot\cdot\frac{n^3-1}{n^3+1}$
But I could'nt find any pattern in this.So how  the limit can be evaluated?

There's some telescoping going on here. We have
$$\frac{n^3-1}{n^3+1}=\frac{n-1}{n+1}\frac{n^2+n+1}{n^2-n+1}.$$
Then
$$\prod_{n=2}^N\frac{n-1}{n+1}=\frac13\frac24\frac35\frac46\cdots\frac{N-1}{N+2}$$
and
$$\prod_{n=2}^N\frac{n^2+n+1}{n^2-n+1}=\frac73\frac{13}7\frac{21}{13}\frac{31}{21}\cdots\frac{N^2+N+1}{N^2-N+1}.$$
Both these products telescope to give a closed form for
$$\prod_{n=2}^N\frac{n^3-1}{n^3+1}.$$
