Limit of a continued product $\prod_{r=3}^n \frac{(r^3+3r)^2}{r^6-64} $ as $n \to \infty$

Question

Evaluate $\lim_{n\to \infty }\prod_{r=3}^n \frac{(r^3+3r)^2}{r^6-64} $

I don't have any ideas on how to approach the problem. Hints or solutions are appreciated.

Answer 1

If you factor the general term, you get:

$$\frac{r^2 \left(r^2+3\right)^2}{(r-2) (r+2) \left(r^2-2 r+4\right) \left(r^2+2 r+4\right)}$$

Notice that $r^2-2r + 4 = (r-1)^2 + 3,\ r^2+2r + 4 = (r+1)^2+3,$ So the product telescopes: and almost everything cancels. (the final answer is $\frac{72}7.)$