International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria


Ivan is Watching You
Ivan's Office

Day 1
    Problem 1
    Problem 2
    Problem 3
    Problem 4
    Problem 5

Day 2
    Problem 6
    Problem 7
    Problem 8
    Problem 9
    Problem 10


    Day 1 questions
    Day 1 solutions
    Day 2 questions
    Day 2 solutions
    Closing Ceremony

Official IMC site

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

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


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

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.