International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria

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

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

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    Day 1 questions
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Problem 1

Problem 1. Let \(\displaystyle (a_n)_{n=1}^{\infty}\) and \(\displaystyle (b_n)_{n=1}^{\infty}\) be two sequences of positive numbers. Show that the following statements are equivalent:

(1) There is a sequence \(\displaystyle (c_n)_{n=1}^{\infty}\) of positive numbers such that \(\displaystyle \displaystyle\sum\limits_{n=1}^{\infty} \dfrac{a_n}{c_n}\) and \(\displaystyle \displaystyle\sum\limits_{n=1}^{\infty} \dfrac{c_n}{b_n}\) both converge;

(2) \(\displaystyle \displaystyle\sum\limits_{n=1}^{\infty} \sqrt{\dfrac{a_n}{b_n}}\) converges.

(Proposed by Tomáš Bárta, Charles University, Prague)

Solution. Note that the sum of a series with positive terms can be either finite or \(\displaystyle +\infty\), so for such a series, "converges" is equivalent to "is finite".

Proof for \(\displaystyle (1)\implies(2)\): By the AM-GM inequality,

\(\displaystyle \sqrt{\dfrac{a_n}{b_n}} = \sqrt{\dfrac{a_n}{c_n}\cdot \dfrac{c_n}{b_n}} \le \frac12\left(\dfrac{a_n}{c_n}+\dfrac{c_n}{b_n}\right), \)

so

\(\displaystyle \sum_{n=1}^\infty\sqrt{\dfrac{a_n}{b_n}} \le \frac12\sum_{n=1}^\infty\dfrac{a_n}{c_n} + \frac12\sum_{n=1}^\infty\dfrac{c_n}{b_n} < +\infty. \)

Hence, \(\displaystyle \displaystyle\sum_{n=1}^\infty\sqrt{\dfrac{a_n}{b_n}}\) is finite and therefore convergent.

Proof for \(\displaystyle (2)\implies(1)\): Choose \(\displaystyle c_n=\sqrt{a_nb_n}\). Then

\(\displaystyle \dfrac{a_n}{c_n} = \dfrac{c_n}{b_n} = \sqrt{\dfrac{a_n}{b_n}}. \)

By the condition \(\displaystyle \displaystyle\sum_{n=1}^\infty\sqrt{\dfrac{a_n}{b_n}}\) converges, therefore \(\displaystyle \displaystyle\sum_{n=1}^\infty\dfrac{a_n}{c_n}\) and \(\displaystyle \displaystyle\sum_{n=1}^\infty\dfrac{c_n}{b_n}\) converge, too.