IMC 2018 · Day 2 · P7

We will estimate the ratio between the terms $|a_{n+2} - a_{n+1}|$ and $|a_{n+1} - a_n|$.

Before doing that, we localize the numbers $a_n$; we prove that $$\tag{2} -2 \le a_n \le -\sqrt[3]{4} \quad \text{for } n \ge 1.$$ The lower bound simply follows from the recurrence: $a_n = \sqrt[3]{a_{n-1}^2 - 8} \ge \sqrt[3]{-8} = -2$. The proof of the upper bound can be done by induction: we have $a_1 = -2 < -\sqrt[3]{4}$, and whenever $-2 \le a_n < 0$, it follows that $a_{n+1} = \sqrt[3]{a_n^2 - 8} \le \sqrt[3]{2^2 - 8} = -\sqrt[3]{4}$.

Now compare $|a_{n+2} - a_{n+1}|$ with $|a_{n+1} - a_n|$. By applying $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$, $x^2 - y^2 = (x - y)(x + y)$ and the recurrence, $$\begin{align*} (a_{n+2}^2 + a_{n+2} a_{n+1} + a_{n+1}^2) \cdot |a_{n+2} - a_{n+1}| &= \\ = |a_{n+2}^3 - a_{n+1}^3| = \bigl| (a_{n+1}^2 - 8) - (a_n^2 - 8) \bigr| &= \\ = |a_{n+1} + a_n| \cdot |a_{n+1} - a_n|. & \end{align*}$$ On the left-hand side we have $$a_{n+2}^2 + a_{n+2} a_{n+1} + a_{n+1}^2 \ge 3 \cdot 4^{2/3};$$ on the right-hand side $$|a_{n+1} + a_n| \le 4.$$ Hence, $$|a_{n+2} - a_{n+1}| \le \frac{4}{3 \cdot 4^{2/3}} |a_{n+1} - a_n| = \frac{\sqrt[3]{4}}{3} |a_{n+1} - a_n|.$$ By a trivial induction it follows that $$|a_{n+1} - a_n| < \left( \frac{\sqrt[3]{4}}{3} \right)^{n-1} |a_2 - a_1|.$$ Hence the series $\sum_{n=0}^{\infty} |a_{n+1} - a_n|$ can be majorized by a geometric series with quotient $\frac{\sqrt[3]{4}}{3} < 1$; that proves that the series converges.