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Let $a_n$,$b_n$ with $b_n>0$ be two bounded sequences which are eventually close to, respectively, two other sequences $\bar a_n$,$\bar b_n$ with $\bar b_n>0$, that is, for every $\epsilon >0$ there exists an integer $N$ such that $|a_n-\bar a_n|<\epsilon$ and $|b_n-\bar b_n|<\epsilon$ for all $n>N$.

Is it possible to prove that the sequence $\frac{a_n}{b_n}$ will be eventually close to the sequence $\frac{\bar a_n}{\bar b_n}$? No assumptions are made on the sequences apart from boundedness and positivity of the denominators, but I'm particularly interested in the case in which all of them converge to zero.

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  • $\begingroup$ Any four sequences converging to zero satisfy your conditions. $\endgroup$ – Brendan McKay Nov 18 '12 at 9:53
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If the sequences are bounded, and the b's are bounded away from 0, there is a simple proof by the triangle equality. If the a's are both 1/n, and the b's are 1/n and -1/n, this clearly fails.

Edit: As far as I can see, this can't be rescued by insisting on positive sequences. Starting with the first counterexample, keep the a's equal to each other, but tending to zero a bit slower than 1/n. Take the b's to be 1/n, and 1/n slightly perturbed, by a factor 1 + o(1). Working out the difference of ratios, it can be made large.

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  • $\begingroup$ Of course you're right; I need to impose that the b's remain positive. But not necessarily bounded away from zero, which is the interesting case. Thanks very much. $\endgroup$ – Roberto López-Valcarce Nov 17 '12 at 14:59
  • $\begingroup$ Well, OK, you don't say why you are interested in the question. For me it is probably related to 2x2 matrices and the geometry of the singular set. $\endgroup$ – Charles Matthews Nov 17 '12 at 15:52
  • $\begingroup$ In fact it came up during the convergence analysis of an iterative algorithm. The numerator and denominator are updated independently, but what is of interest is actually the ratio, wich seems to converge to the desired value even if both a's and b's tend to zero. $\endgroup$ – Roberto López-Valcarce Nov 17 '12 at 16:17
  • $\begingroup$ I guess you're right. Any two sequences converging to the same limit, and to zero in particular, are "eventually close" to each other as per the definition above, regardless of their individual convergence rates. This allows to construct a myriad counterexamples by properly choosing these rates in which the ratios have different limits. Thank you. $\endgroup$ – Roberto López-Valcarce Nov 17 '12 at 21:11

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