<s>[INST] Which number is more near to $S$ Which number is more near to $S$?
$S=\frac{(2^3-1)(3^3-1)\dots (100^3-1)}{(2^3+1)(3^3+1)\dots (100^3+1)}$

1.$0.6$
2.$0.67$
3.$0.667$
4.$0.6667$

I personaly can not do anything special I have tried to calculate some easy case but no resaults any hints?
 [/INST] \begin{align*}
  \prod_{k=2}^{n} \frac{k^3-1}{k^3+1} &=
  \prod_{k=2}^{n} \frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)} \\
  &= \prod_{k=2}^{n} \frac{k-1}{k+1} \times \frac{k^2+k+1}{(k-1)^2+(k-1)+1} \\
  &= \left(
       \frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \frac{4}{6}  \cdot
       \ldots \cdot \frac{n-3}{n-1} \cdot \frac{n-2}{n} \cdot \frac{n-1}{n+1}
     \right) \times \frac{n^2+n+1}{3} \\
  &= \frac{2(n^2+n+1)}{3n(n+1)} \\
 S &= \frac{2(100^2+100+1)}{3(100\times 101)} \\
   &= \frac{2}{3}\left(1+\frac{1}{10100} \right) \\
   &\approx 0.6667
\end{align*}
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