Let $\{a_k\}$ and $\{b_k\}$ be positive sequences in $\mathbb{R}$ that both converge to zero. Can we choose $\{c_k\}$ such that it converges to zero and

$$ 0<\lim_{k \to \infty} \frac{a_k}{c_k} = \lim_{k \to \infty} \frac{b_k}{c_k} < +\infty $$

  • $\begingroup$ $a_k$ and $b_k$ are given. We can only decide on $c_k$ $\endgroup$ – Mehdi Jafarnia Jahromi Apr 27 '15 at 23:14
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    $\begingroup$ I think that would be very difficult for $a_k=e^{-k}$ and $b_k=1/k$. I am pretty certain you would need $\lim_{k\to\infty}a_k/b_k=1$ $\endgroup$ – Arthur Apr 27 '15 at 23:14
  • $\begingroup$ @arthur: Actually, not that hard. Just take c_k that goes slower to 0 than both a_k and b_k. Like 1/ln(k). Both limits will then be +oo (But yes, I'm not sure that's what the OP had in head) $\endgroup$ – Tryss Apr 27 '15 at 23:19
  • $\begingroup$ @Tryss As long as you think $\infty=\infty$ is a correct statement, then you're right (except that you would want $c_k$ to go to zero faster, not slower). I know many would disagree. $\endgroup$ – Arthur Apr 27 '15 at 23:22
  • $\begingroup$ I edited the question, we do not need $+\infty$ $\endgroup$ – Mehdi Jafarnia Jahromi Apr 27 '15 at 23:24

This is not possible in general, for example, consider the sequences $(a_n) = (\frac{1}{n})$ and $(b_n)=(\frac{1}{n^2})$. Then if possible let $(c_k)$ be a sequence such that \begin{equation} 0<\lim_\limits{n\to \infty}\frac{a_n}{c_n} =\lim_\limits{n\to \infty}\frac{b_n}{c_n}=l<\infty. \end{equation} Then it will follow that \begin{equation} \frac{\lim_\limits{n\to \infty}\frac{a_n}{c_n}}{\lim_\limits{n\to \infty}\frac{b_n}{c_n}} =1= \lim_\limits{n\to \infty}\frac{a_n}{b_n}=\lim_\limits{n\to \infty}n. \end{equation} which is obviously not true.

$\textbf{Note:}$ The above observation shows that it is crucial to have $\lim_\limits{n\to \infty}\frac{a_n}{b_n}=1$ in the hypothesis and if this is the case then it can be easily seen that taking the sequence $(c_n) = (b_n)$ will do the job.

$\textbf{Observation}:$ Suppose if it is added in the hypothesis that $\lim_\limits{n\to \infty}\frac{a_n}{b_n}=1$ and it is asked that whether there is a sequence $(c_n)$ such that $(c_n)$ converges to $0$ and \begin{equation} 0<\lim_\limits{n\to \infty}\frac{a_n}{c_n} =\lim_\limits{n\to \infty}\frac{b_n}{c_n}=l<\infty. \end{equation} and $l$ is different than $1$. Even in this case the answer turns out to be affirmative. Suppose $l\in\mathbb{R}$ is any real number different from $1$. Take the sequence $(c_n) =(\frac{a_n^2}{lb_n})$. Then it is clear that $c_n\longrightarrow 0$ and moreover, \begin{equation} 0<\lim_\limits{n\to \infty}\frac{a_n}{c_n} =\lim_\limits{n\to \infty}\frac{b_n}{c_n}=l<\infty. \end{equation}


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