The answer is $$2/3$$

The trick is to know the simplification:

$$
\frac{T^3-1}{T^3+1} = \left( \frac{T-1}{T+1} \right) \left( \frac{T^2+T+1}{T^2-T+1} \right)
$$

This lets us prove by induction that:

$$
\prod_{n=2}^{T} \frac{n^3-1}{n^3+1} = \frac{2}{3} \left( \frac{T^2+T+1}{T^2+T} \right)
$$

which becomes $$2/3$$ in the limit as $$T \rightarrow \infty$$

To prove it, we can verify the equation easily for $$T=2$$.

Then:

$$
\prod_{n=2}^{T} \frac{n^3-1}{n^3+1} = \left( \prod_{n=2}^{T-1} \frac{n^3-1}{n^3+1} \right) \left( \frac{T^3-1}{T^3+1} \right)
$$

$$
= \frac{2}{3} \left( \frac{(T-1)^2 + (T-1) + 1}{(T-1)^2 + (T-1)} \right) \left( \frac{T-1}{T+1} \right) \left( \frac{T^2+T+1}{T^2-T+1} \right)
$$

$$
= \frac{2}{3} \left( \frac{T^2 - T + 1}{T^2-T} \right) \left( \frac{T-1}{T+1} \right) \left( \frac{T^2+T+1}{T^2-T+1} \right)
$$

$$
= \frac{2}{3} \left( \frac{T^2+T+1}{T^2+T} \right)
$$

Q.E.D.

SimpleStacker

science

Nothing too complicated to get started again.

Solve this infinite product:
$$
\prod_{n=2}^{\infty} \frac{n^3-1}{n^3+1}=\ ?
$$

Previous iteration: https://stacker.news/items/770290/r/south_korea_ln