IMC 2019 · Day 1 · P1

Hint: Telescoping product.

Let $$a_n = \frac{(n^3 + 3n)^2}{n^6 - 64}.$$ Notice that $$\begin{align*} a_n &= \frac{(n^3 + 3n)^2}{(n^3 - 8)(n^3 + 8)} = \frac{n^2 (n^2 + 3)^2} {(n - 2)(n^2 + 2n + 4) \cdot (n + 2)(n^2 - 2n + 4)} \\ &= \frac{n}{n-2} \cdot \frac{n}{n+2} \cdot \frac{n^2 + 3}{(n-1)^2 + 3} \cdot \frac{n^2 + 3}{(n+1)^2 + 3}. \end{align*}$$ Hence, for $N \ge 3$ we have $$\begin{align*} \prod_{n=3}^{N} a_n &= \left( \prod_{n=3}^{N} \frac{n}{n-2} \right) \left( \prod_{n=3}^{N} \frac{n}{n+2} \right) \left( \prod_{n=3}^{N} \frac{n^2 + 3}{(n-1)^2 + 3} \right) \left( \prod_{n=3}^{N} \frac{n^2 + 3}{(n+1)^2 + 3} \right) \\ &= \frac{N (N-1)}{1 \cdot 2} \cdot \frac{3 \cdot 4}{(N+1)(N+2)} \cdot \frac{N^2 + 3}{2^2 + 3} \cdot \frac{3^2 + 3}{(N+1)^2 + 3} \\ &= \frac{72}{7} \cdot \frac{N (N-1)(N^2 + 3)} {(N+1)(N+2) \bigl( (N+1)^2 + 3 \bigr)} \\ &= \frac{72}{7} \cdot \frac{(1 - \frac1N)(1 + \frac{3}{N^2})} {(1 + \frac1N)(1 + \frac2N) \bigl( (1 + \frac1N)^2 + \frac{3}{N^2} \bigr)}, \end{align*}$$ so $$\prod_{n=3}^{\infty} a_n = \lim_{N \to \infty} \prod_{n=3}^{N} a_n = \lim_{N \to \infty} \left( \frac{72}{7} \cdot \frac{(1 - \frac1N)(1 + \frac{3}{N^2})} {(1 + \frac1N)(1 + \frac2N) \bigl( (1 + \frac1N)^2 + \frac{3}{N^2} \bigr)} \right) = \frac{72}{7}.$$