### International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria

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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.