hi
Let
P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]
n=2,3,4,......
Find Limit P when N tends to infinity.
You easily undestand if you write the sequence of terms...
$\displaystyle P= \prod_{n=2}^{\infty} \frac{(n-1)\cdot (n^{2} + n + 1)}{(n+1)\cdot (n^{2} - n + 1)}= \frac {1\cdot 7}{3\cdot 3}\cdot \frac {2\cdot 13}{4\cdot 7}\cdot \frac {3\cdot 21}{5\cdot 13}\cdot \frac {4\cdot 31}{6\cdot 21}\cdot \dots$
... and it is evident you can simplify both in numerator and denominator [3 with 3, 4 with 4, ... , 7 with 7, 13 with 13, 21 with 21,...]
Finally is...
$\displaystyle P=\frac{2}{3}$
Very nice!...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$