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Infinite Product

Evaluate the infinite product:

    \[\Pi = \prod_{n=3}^{\infty} \frac{\left ( n^3+3n \right )^2}{n^6-64}\]

(IMC 2019 / Day 1 / Problem 1)

Solution

    \begin{align*} \prod_{n=3}^{\infty} \frac{\left ( n^3+3n \right )^2}{n^6-64} &= \prod_{n=3}^{\infty} \frac{n^2 \left ( n^2+3 \right )^2}{n^6-69} \\ &=\prod_{n=3}^{\infty} \frac{n^2\left ( n^2+3 \right )^2}{\left ( n^3-8 \right )\left ( n^3+8 \right )} \\ &= \prod_{n=3}^{\infty} \frac{n^2\left ( n^2+3 \right )^2}{\left ( n-2 \right )\left ( n^2+2n+4 \right )\left ( n+2 \right )\left ( n^2-2n+4 \right )} \\ &=\prod_{n=3}^{\infty} \frac{n}{n-2} \cdot \frac{n}{n+2} \cdot \frac{n^2+3}{\left ( n-1 \right )^2+3} \cdot \frac{n^2+3}{\left ( n+1 \right )^2+3} \\ &= \lim_{N \rightarrow +\infty} \frac{N\left ( N-1 \right )}{1 \cdot 2} \cdot \frac{3 \cdot 4}{\left ( N+1 \right ) \left ( N+2 \right )} \cdot \frac{N^2+3}{2^2+3} \cdot \\ & \quad \quad \quad \quad \quad \cdot\frac{3^2+3}{\left ( N+1 \right )^2+3} \\ &= \frac{72}{7} \end{align*}

since the product telescopes.

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