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How to find a sequence having the following properties:

$\dfrac{A_n}{B_n} \to 0$, but $\dfrac{\log(A_n)}{\log(B_n)} \to 1$ as $n \to \infty$.

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Sanch, I see that this is your first question. So I wanted to let you know a few things about MathSE. We like to know the sources of questions - if it's homework, add the [homework] tag. People will still help, so don't worry. We also like to know what you've tried on a problem. These sort of pleasantries usually result in more and better answers. –  mixedmath Sep 3 '11 at 19:17
    
@mixedmath , Thanks a lot for your advise. I will keep this in mind :) –  Sanch Sep 3 '11 at 20:02
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For $n \ge 2$, let $a_n=n$, and $b_n=n \log n$. –  André Nicolas Sep 3 '11 at 20:14

2 Answers 2

up vote 4 down vote accepted

Let $A_n = \exp n^2$, $B_n = \exp (n^2 + n)$.

$$ A_n /B_n = \exp (-n) \to 0 $$

while

$$ \ln A_n / \ln B_n = n^2 / (n^2 + n) = 1 / (1 + n^{-1}) \to 1$$


Here's how I arrived at the example, the conditions you wrote down requires that

$$ \log_{B_n} A_n = 1 - \epsilon(n) $$

where $\epsilon(n) \to 0$. So we write $A_n = (B_n)^{1-\epsilon(n)}$, and realise that the first condition on $A_n/B_n$ requires that $B_n^{-\epsilon(n)} \to 0$, or $\exp (-\epsilon(n) \ln B_n) \to 0$. This means that $\epsilon(n) \ln B_n$ must diverge to infinity, which would be the case if $\ln B_n = \epsilon(n)^{-1 - \delta}$ for any $\delta > 0$.

So if we take any positive function $\epsilon(n)$ with the property that $\lim_{n\to \infty} \epsilon(n) = 0$, and any $\delta > 0$, the sequences

$$ B_n = \exp ( \epsilon(n)^{-1 - \delta} ) $$

and

$$ A_n = B_n^{1 - \epsilon(n)} = \exp ( \epsilon(n)^{-1 - \delta} - \epsilon(n)^{-\delta} ) $$

would satisfy the requirement. The example I wrote down is, morally speaking, for the case $\delta = 1$, $\epsilon(n) = n^{-1}$.

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Thanks a lot for your help :) –  Sanch Sep 3 '11 at 20:23

Well, if $a_i = 2^n$ and $b_i = 3^n$, then $\dfrac{\log a_i}{\log b_i} = \dfrac{2}{3}$. Does that give you the inspiration to come up with such a sequence?

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I had tried this. But couldn't come up with the proper answer. –  Sanch Sep 3 '11 at 20:24

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