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IMC 2019 1

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\begin{align} \prod_{n=3}^{\infty} \frac{(n^3+3n)^2}{n^6-64} &= \prod_{n=3}^{\infty} \frac{n^2(n^2+3)^2}{n^6-64} \\ &= \prod_{n=3}^{\infty} \frac{n^2(n^2+3)^2}{(n^2-4)(n^4+4n^2+16)} \\ &= \left(\prod_{n=3}^{\infty} \frac{n^2}{n^2-4}\right)\left(\prod_{n=3}^{\infty} \frac{(n^2+3)^2}{n^4+4n^2+16}\right) \end{align}

consider (to find its roots and thus factorise it)

\begin{align} n^4+4n^2+16&=0 \\ \iff n^2&=\frac{-4±\sqrt{16-64}}{2} \\ &= -2±\sqrt{-12} \\ &= -2±2i\sqrt3 \\ &= 4e^{i\pi(±\frac 2 3 +2k)} \ \forall \ k\in\mathbb{Z} \\ \iff n&=(4e^{i\pi(±\frac 2 3 +2k)})^{\frac 1 2 } \\ &= 2e^{i\pi(±\frac 1 3 +k)} \end{align}

using de moivre’s theorem. we have a quartic so there are 4 roots (2 pairs of complex conjugates).
we get that the roots are \displaystyle n=2e^{±\frac{i\pi}{3}},2e^{±\frac{5i\pi}{3}}

\begin{align} \therefore n^4+4n^2+16 &=(n-2e^{\frac{i\pi}{3}})(n-2e^{-\frac{i\pi}{3}})(n-2e^{\frac{5i\pi}{3}})(n-2e^{-\frac{5i\pi}{3}}) \\ &= (n^2-2(e^{\frac{i\pi}{3}}+e^{-\frac{i\pi}{3}})n+4e^{\frac{i\pi}{3}}e^{-\frac{i\pi}{3}})(n^2-2(e^{\frac{5i\pi}{3}}+e^{-\frac{5i\pi}{3}})n+4e^{\frac{5i\pi}{3}}e^{-\frac{5i\pi}{3}}) \\ &= (n^2-4n\cos\frac{\pi}{3}+4)(n^2-4n\cos\frac{5\pi}{3}+4) \\ &= (n^2-2n+4)(n^2+2n+4) \end{align}

alternatively (just spotted this much simpler alternative, still complex method was a nice alternative i think, although would be more useful in other situations), just note that

\begin{align} n^4+4n^2+16 &= n^4+8n^2+16-4n^2 \\ &= (n^2+4)^2-(2n)^2 \\ &= (n^2-2n+4)(n^2+2n+4) \end{align}

now we have

\begin{align} \left(\prod_{n=3}^{\infty} \frac{n^2}{n^2-4}\right)\left(\prod_{n=3}^{\infty} \frac{(n^2+3)^2}{n^4+4n^2+16}\right) &= \left(\prod_{n=3}^{\infty} \frac{n}{n-2}\right)\left(\prod_{n=3}^{\infty} \frac{n}{n+2}\right)\left(\prod_{n=3}^{\infty} \frac{(n^2+3)^2}{(n^2-2n+4)(n^2+2n+4)}\right) \\ &= \left(\prod_{n=3}^{\infty} \frac{n}{n-2}\right)\left(\prod_{n=3}^{\infty} \frac{n}{n+2}\right)\left(\prod_{n=3}^{\infty} \frac{n^2+3}{n^2-2n+4}\right)\left(\prod_{n=3}^{\infty} \frac{n^2+3}{n^2+2n+4}\right) \\ &=\left(\prod_{n=3}^{\infty} \frac{n}{n-2}\right)\left(\prod_{n=3}^{\infty} \frac{n}{n+2}\right)\left(\prod_{n=3}^{\infty} \frac{n^2+3}{(n-1)^2+3}\right)\left(\prod_{n=3}^{\infty} \frac{n^2+3}{(n+1)^2+3}\right) \\ &=\left(\lim_{N\to\infty}\prod_{n=3}^{N}n\right) \left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{n-2}\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N}n\right) \left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{n+2}\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N}n^2+3\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{(n-1)^2+3}\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N}n^2+3\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{(n+1)^2+3}\right) \\ &= \left(\lim_{N\to\infty}\prod_{n=5}^{N+2}n-2\right) \left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{n-2}\right)\left(\lim_{N\to\infty}\prod_{n=1}^{N-2}n+2\right) \left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{n+2}\right)\left(\lim_{N\to\infty}\prod_{n=4}^{N+1}(n-1)^2+3\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{(n-1)^2+3}\right)\left(\lim_{N\to\infty}\prod_{n=2}^{N-1}(n+1)^2+3\right)\left(\lim_{N\to\infty}\prod_{n=3}^{N} \frac{1}{(n+1)^2+3}\right) \\ &= \left(\lim_{N\to\infty}\frac 1 2 N(N-1)\right)\left(\lim_{N\to\infty}\frac{3\times4}{(N+1)(N+2)}\right)\left(\lim_{N\to\infty}\frac{N^2+3}{7}\right)\left(\lim_{N\to\infty}\frac{12}{(N+1)^2+3}\right) \\ &= \lim_{N\to\infty} \frac{N\times(N-1)\times3\times4\times(N^2+3)\times12}{2\times (N+1)\times(N+2)\times7\times((N+1)^2+3)} \end{align}

as N\to\infty, N+a\to N where a\in\mathbb{R}. so we can simplify this to:

\begin{align} \lim_{N\to\infty} \frac{N\times(N-1)\times3\times4\times(N^2+3)\times12}{2\times (N+1)\times(N+2)\times7\times((N+1)^2+3)} &=\frac{\cancel{N}\times \cancel{N}\times3\times4\times\cancel{(N^2+3)}\times12}{2\times \cancel{N}\times \cancel{N}\times7\times\cancel{(N^2+3)}} \\ &= \frac{72}7 \end{align}
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