<s>[INST] Limit of a continued product $\prod_{r=3}^n \frac{(r^3+3r)^2}{r^6-64} $ as $n \to \infty$ 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.
 [/INST] 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 $\frac{72}7.)$
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