2019 IMC - Telescoping series

$\begin{aligned} \prod_{n=3}^\infty \frac {(n^3+3n)^2}{n^6-64} & = \prod_{n=3}^\infty \frac {n \cdot n (n^2+3)(n^2+3)}{(n-2)(n+2)(n^2-2n+4)(n^2+2n+4)} \\ & = \prod_{n=3}^\infty \frac {n \cdot n (n^2+3)(n^2+3)}{(n-2)(n+2)((n-1)^2+3)((n+1)^2+3)} \\ & = \prod_\blue{n=3}^\infty \frac n{n-2} \cdot \prod_{n=3}^\infty \frac n{n+2} \cdot \prod_\blue{n=3}^\infty \frac {n^2+3}{(n-1)^2+3} \cdot \prod_{n=3}^\infty \frac {n^2+3}{(n+1)^2+3} \\ & = \prod_\red{n=1}^\infty \frac {n+2}n \cdot \prod_{n=3}^\infty \frac n{n+2} \cdot \prod_\red{n=2}^\infty \frac {(n+1)^2+3}{n^2+3} \cdot \prod_{n=3}^\infty \frac {n^2+3}{(n+1)^2+3} \\ & = \frac {3 \cdot 4}{1 \cdot 2} \prod_\blue{n=3}^\infty \left(\cancel{\frac {n+2}n \cdot \frac n{n+2}} ^1\right) \cdot \frac {12}7 \prod_\blue{n=3}^\infty \left(\cancel{\frac {(n+1)^2+3}{n^2+3} \cdot \frac {n^2+3}{(n+1)^2+3}}^1 \right) \\ & = \boxed{\frac {72}7} \end{aligned}$

Define the sequence $\{x_n\}$ as, $x_n=\frac{(n^3+3n)^2}{n^6-64}$ The idea is to decompose $x_n$ into simple fractions that might telescope. So we rewrite the denominator of the general term, as $\begin{aligned} n^6-64&= (n^3+8)(n^3-8) \\ & =(n+2)(n^2-2n+4)(n-2)(n^2+2n+4) \\ &=(n+2)\cdot \left((n-1)^2+3\right)\cdot (n-2)\cdot \left((n+1)^2+3\right) \end{aligned}$ Since the denominator is a product of four terms, we shall try doing the same for the numerator. $\begin{aligned} (n^3+3n)^2&=\left(n(n^2+3)\right)^2 \\ &=n^2(n^2+3)^2 \\ &=n\cdot n\cdot (n^2+3)\cdot (n^2+3) \end{aligned}$ So, the general term of the sequence is, $\begin{aligned} x_n=\frac{(n^3+3n)^2}{n^6-64} &=\frac{n\cdot n\cdot (n^2+3)\cdot (n^2+3)}{(n+2)\cdot \left((n-1)^2+3\right)\cdot (n-2)\cdot \left((n+1)^2+3\right)} \\ \\ &=\left(\frac{n}{n+2}\right)\cdot \left(\frac{n}{n-2}\right)\cdot \left(\frac{n^2+3}{(n+1)^2+3}\right)\cdot \left(\frac{n^2+3}{(n-1)^2+3}\right) \end{aligned}$ We shall use the following infinite products: $\begin{aligned} &\prod_{n=3}^{\infty} \frac{n}{n+2}=\lim_{k\to \infty}\prod_{n=3}^{k} \frac{n}{n+2} =\lim_{k\to \infty} \left(\frac{3}{5}\times \frac{4}{6}\times\frac{5}{7}\times\frac{6}{8}\times \cdots \times \frac{k-1}{k+1} \times \frac{k}{k+2}\right) =\lim_{k\to \infty } \frac{12}{(k+1)(k+2)} \\ \\ &\prod_{n=3}^{\infty} \frac{n}{n-2}=\lim_{k\to \infty}\prod_{n=3}^{k} \frac{n}{n-2} =\lim_{k\to \infty} \left(\frac{3}{1}\times \frac{4}{2}\times\frac{5}{3}\times\frac{6}{4}\times \cdots \times \frac{k-1}{k-3} \times \frac{k}{k-2}\right) =\lim_{k\to \infty } \frac{(k-1)k}{2} \\ \\ &\prod_{n=3}^{\infty} \frac{n^2+3}{(n+1)^2+3}= \lim_{k\to \infty}\prod_{n=3}^{k} \frac{n^2+3}{(n+1)^2+3} =\lim_{k\to \infty} \left(\frac{12}{19}\times \frac{19}{28}\times\frac{28}{39}\times\cdots \times \frac{(k-1)^2+3}{k^2+3} \times \frac{k^2+3}{(k+1)^2+3}\right) =\lim_{k\to \infty} \frac{12}{(k+1)^2+3} \\ \\ &\prod_{n=3}^{\infty} \frac{n^2+3}{(n-1)^2+3}= \lim_{k\to \infty}\prod_{n=3}^{k} \frac{n^2+3}{(n-1)^2+3} =\lim_{k\to \infty} \left(\frac{12}{7}\times \frac{19}{12}\times\frac{28}{19}\times\cdots \times \frac{(k-1)^2+3}{(k-2)^2+3} \times \frac{k^2+3}{(k-1)^2+3}\right) =\lim_{k\to \infty} \frac{k^2+3}{7} \end{aligned}$ Therefore, the product of all terms of the sequence $\{x_n\}$ is, $\begin{aligned} \prod_{n=3}^{\infty} x_n &=\left(\prod_{n=3}^{\infty}\frac{n}{n+2}\right)\cdot \left(\prod_{n=3}^{\infty} \frac{n}{n-2} \right)\cdot \left(\prod_{n=3}^{\infty} \frac{n^2+3}{(n+1)^2+3}\right)\cdot \left(\prod_{n=3}^{\infty} \frac{n^2+3}{(n-1)^2+3}\right) \\ \\ &=\lim_{k\to \infty} \left( \frac{12}{(k+1)(k+2)}\cdot \frac{(k-1)k}{2}\cdot \frac{12}{(k+1)^2+3}\cdot \frac{k^2+3}{7}\right) \\ \\ &=\lim_{k\to \infty} \frac{72(k^2-1)(k^2+3)}{7(k^2+3k+2)(k^2+2k+4)} \\ \\ &=\frac{72}{7}\cdot \lim_{k\to \infty} \frac{\left(1-\dfrac{1}{k^2}\right) \left(1+\dfrac{3}{k^2}\right)}{\left(1+\dfrac{3}{k}+\dfrac{2}{k^2}\right)\left(1+\dfrac{2}{k}+\dfrac{4}{k^2}\right)} \\ &=\boxed{\frac{72}{7}} \end{aligned}$