International Mathematics Competition
for University Students
2019

IMC2019: Day 1, Problem 1

Problem 1. Evaluate the product

$\displaystyle \prod_{n=3}^{\infty}\frac{(n^3+3n)^2}{n^6-64} .$

Proposed by Orif Ibrogimov, ETH Zurich and National University of Uzbekistan and

Karen Keryan, Yerevan State University and American University of Armenia, Yerevan

Solution. Let

$\displaystyle a_n = \frac{(n^3+3n)^2}{n^6-64}.$

Notice that

$\displaystyle \begin{align*} a_n & = \frac{(n^3+3n)^2}{(n^3-8)(n^3+8)} = \frac{n^2(n^2+3)^2}{(n-2)(n^2+2n+4)\cdot(n+2)(n^2-2n+4)} \\ & = \frac{n}{n-2} \cdot \frac{n}{n+2} \cdot \frac{n^2+3}{(n-1)^2+3} \cdot \frac{n^2+3}{(n+1)^2+3}. \end{align*}$

Hence, for $\displaystyle N\ge3$ we have

$\displaystyle \begin{align*} \prod_{n=3}^{N} a_n & = \left(\prod_{n=3}^{N} \frac{n}{n-2} \right) \left(\prod_{n=3}^{N} \frac{n}{n+2} \right) \left(\prod_{n=3}^{N} \frac{n^2+3}{(n-1)^2+3} \right) \left(\prod_{n=3}^{N} \frac{n^2+3}{(n+1)^2+3} \right) \\ & = \frac{N(N-1)}{1\cdot2} \cdot \frac{3\cdot 4}{(N+1)(N+2)} \cdot \frac{N^2+3}{2^2+3} \cdot \frac{3^2+3}{(N+1)^2+3} \\ & = \frac{72}{7} \cdot \frac{N(N-1)(N^2+3)}{(N+1)(N+2)\big((N+1)^2+3\big)} \\ & = \frac{72}{7} \cdot \frac{(1-\tfrac1N)(1+\tfrac{3}{N^2})}{(1+\tfrac1N)(1+\tfrac2N)\big((1+\tfrac1N)^2+\tfrac3{N^2}\big)}, \end{align*}$

so

$\displaystyle \prod_{n=3}^{\infty} a_n = \lim_{N\to\infty} \prod_{n=3}^{N} a_n = \lim_{N\to\infty} \left(\frac{72}{7} \cdot \frac{(1-\tfrac1N)(1+\tfrac{3}{N^2})}{(1+\tfrac1N)(1+\tfrac2N)\big((1+\tfrac1N)^2+\tfrac3{N^2}\big)}\right) = \frac{72}{7}.$