IMC 2010 · Day 1 · P3

Let $y_n = x_n^2$. Then $y_{n+1} = (y_n - 2)^2$ and $y_{n+1} - 4 = y_n (y_n - 4)$. Since $y_2 = 9 > 5$, we have $y_3 = (y_2 - 2)^2 > 5$ and inductively $y_n > 5$, $n \ge 2$. Hence, $y_{n+1} - y_n = y_n^2 - 5 y_n + 4 > 4$ for all $n \ge 2$, so $y_n \to \infty$.

By $y_{n+1} - 4 = y_n (y_n - 4)$, $$\begin{align*} \left( \frac{x_1 \cdot x_2 \cdot x_3 \cdots x_n}{x_{n+1}} \right)^2 &= \frac{y_1 \cdot y_2 \cdot y_3 \cdots y_n}{y_{n+1}} \\ &= \frac{y_{n+1} - 4}{y_{n+1}} \cdot \frac{y_1 \cdot y_2 \cdot y_3 \cdots y_n}{y_n - 4} = \frac{y_{n+1} - 4}{y_{n+1}} \cdot \frac{y_1 \cdot y_2 \cdot y_3 \cdots y_{n-1}}{y_n - 4} \cdot y_n = \cdots = \\$$

&= \frac{y_{n+1} - 4}{y_{n+1}} \cdot \frac{1}{y_1 - 4} = \frac{y_{n+1} - 4}{y_{n+1}} \to 1. \end{align*} Therefore, $$\lim_{n \to \infty} \frac{x_1 \cdot x_2 \cdot x_3 \cdots x_n}{x_{n+1}} = 1.$$