### International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria

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### Problem 7

Problem 7. Let $\displaystyle (a_n)_{n=0}^\infty$ be a sequence of real numbers such that $\displaystyle a_0=0$ and

$\displaystyle a_{n+1}^3=a_n^2-8 \quad \text{for} \quad n=0,1,2,\ldots$

Prove that the following series is convergent:

 $\displaystyle \sum_{n=0}^\infty|a_{n+1}-a_n|.$ $\displaystyle (1)$

(Proposed by Orif Ibrogimov, National University of Uzbekistan)

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

Before doing that, we localize the numbers $\displaystyle a_n$; we prove that

 $\displaystyle -2 \le a_n \le -\sqrt[3]{4} \quad\text{for $$\displaystyle n\ge1$.}$$ $\displaystyle (2)$

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

Now compare $\displaystyle |a_{n+2}-a_{n+1}|$ with $\displaystyle |a_{n+1}-a_{n}|$. By applying $\displaystyle {x^3-y^3=(x-y)(x^2+xy+y^2)}$, $\displaystyle {x^2-y^2=(x-y)(x+y)}$ and the recurrence,

\displaystyle \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| &= \big|(a_{n+1}^2-8)-(a_{n}^2-8)\big| = \\ &= |a_{n+1}+a_{n}| \cdot |a_{n+1}-a_{n}|. \end{align*}

On the left-hand side we have

$\displaystyle a_{n+2}^2+a_{n+2}a_{n+1}+a_{n+1}^2 \ge 3\cdot 4^{2/3};$

on the right-hand side

$\displaystyle |a_{n+1}+a_{n}| \le 4.$

Hence,

$\displaystyle |a_{n+2}-a_{n+1}| \leq \dfrac{4}{3\cdot 4^{2/3}} |a_{n+1}-a_n| = \dfrac{\sqrt[3]{4}}{3} |a_{n+1}-a_n|.$

By a trivial induction it follows that

$\displaystyle |a_{n+1}-a_n| < \left(\dfrac{\sqrt[3]{4}}{3}\right)^{n-1}|a_2-a_1|.$

Hence the series $\displaystyle \sum_{n=0}^\infty|a_{n+1}-a_n|$ can be majorized by a geometric series with quotient $\displaystyle \dfrac{\sqrt[3]{4}}{3}<1$; that proves that the series converges.