IMC 2018 P1

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#1 h Jul 24, 2018, 4:33 PM • 4 Y

Y by Davi-8191, mathematicsy, Adventure10, Mango247

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

There is a sequence $(c_n)_{n=1}^{\infty}$ of positive numbers such that $\sum_{n=1}^{\infty}{\frac{a_n}{c_n}}$ and $\sum_{n=1}^{\infty}{\frac{c_n}{b_n}}$ both converge;
$\sum_{n=1}^{\infty}{\sqrt{\frac{a_n}{b_n}}}$ converges.

Proposed by Tomáš Bárta, Charles University, Prague

This post has been edited 1 time. Last edited by ThE-dArK-lOrD, Jul 24, 2018, 4:35 PM

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#2 h Jul 24, 2018, 5:08 PM • 2 Y

Y by Adventure10, Mango247

For $1\implies 2$ ; note that, using AM-GM, we have, $\frac{a_n}{c_n}+\frac{c_n}{b_n}\geq 2\sqrt{\frac{a_n}{b_n}}$ . Hence,
$2\sum_n \sqrt{\frac{a_n}{b_n}} \leq \sum_n \frac{a_n}{c_n}+\sum_n \frac{c_n}{b_n}.$
For $2\implies 1$ , just take $c_n=\sqrt{a_nb_n}$ , and $a_n/c_n = \sqrt{a_n/b_n}=c_n/b_n$ . Thus, both converge.

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#3 h Jun 21, 2020, 7:59 PM

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$\text{Indeed AM-GM is the key}$
$\text{Lemma: If}$ $S=\sum_{k=1}^{\infty}x_n$ $,S'=\sum_{k=1}^{\infty}y_n$ $\text{both converges then}$ $\sum_{k=1}^{\infty} (x_n +y_n)$ $\text{also converges}$
$\text{Proof: If S and S' converges then the sequence of partial sums}$ $S_n,S'_n$ $\text{also converges, where}$ $S_n= \sum_{k=1}^{\infty}x_n$ $S'_n= \sum_{k=1}^{\infty} y_n$ $\text{So}$ $\forall \epsilon> 0 ~ \in \mathbb{R}~\exists N_1~\in \mathbb{N}~$ $\text{such that}$ $\forall n>N_1~ \|S_n-S|<\frac{\epsilon}{2}$ $\text{Similarly}$ $\forall \epsilon> 0 ~ \in \mathbb{R}~\exists N_2~\in \mathbb{N}~$ $\text{such that}$ $\forall n>N_2~ \|S'_n-S|<\frac{\epsilon}{2}$ $\text{Let}$ $N=max(N_1,N_2)$ $\text{Thus by triangle inequality by summing the inequalities}$ $|S_n+S'_n-(S+S')|<\epsilon ~\forall N >n \Rightarrow \sum_{k=1}^{\infty} x_n + \sum_{k=1}^{\infty} y_n \to S+S'$ $\text{Since terms of sequence are positive the sum is invariant under rearrangement and we take the desired rearrangement}$
$\text{So by lemma}$ $\sum_n (a_n/c_n + c_n/b_n)$ $\text{Converges}$ and $\sum_n (a_n/c_n + c_n/b_n)>2\sum_n \sqrt{\frac{a_n}{b_n}}$
$\text{Hence we conclude by Direct Comparison Test}$
$\text{For 2 do as @grupyorum said}$

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#4 h Apr 13, 2025, 8:42 AM

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First let's show $1\implies 2$ .
Let $\sum_{n=1}^{\infty} \frac{a_n}{c_n}$ converges to $a$ and $\sum_{n=1}^{\infty} \frac{c_n}{b_n}$ converges to $b$ .
Then by Cauchy-Schwarz inequality, we have
$\sum_{n=1}^k \sqrt{\frac{a_n}{b_n}}\le \sqrt{\left(\sum_{n=1}^{k} \frac{a_n}{c_n}\right)\left(\sum_{n=1}^{k} \frac{c_n}{b_n}\right)}< \sqrt{\left(\sum_{n=1}^{\infty} \frac{a_n}{c_n}\right)\left(\sum_{n=1}^{\infty} \frac{c_n}{b_n}\right)}= \sqrt{ab}$ [Note that we can do this because $a_n/c_n$ and $c_n/b_n$ are positive numbers, so the sequence of partial sums are increasing.]
So the sequence of partial sums of the sequence $\sqrt{a_n/b_n}$ is bounded above and is also monotonically increasing (since $\sqrt{a_n/b_n}$ is increasing). Hence the series must converge (by monotone convergence theorem).

Now let's show $2\implies 1$ .
Just take $c_n=\sqrt{a_nb_n}$ . So, we have $\sum_{n=1}^{\infty} a_n/c_n = \sum_{n=1}^{\infty} c_n/b_n = \sum_{n=1}^{\infty} \sqrt{a_nb_n}$ which converges.

This post has been edited 1 time. Last edited by Fibonacci_math, Apr 13, 2025, 8:47 AM

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