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I would like to know the sign of the following term in general. I tried approximation for $\log$ function and it had negative sign. Is there any $m_0$ such that for all $n>m>m_0$, the following function is positive or as I get it is always negative in its domain.

$$ f(m,n)=\frac{\log\log\log m}{\log\log\log n} \centerdot \frac{\log\log n}{\log\log m}-\left(1+\frac{\log(\frac n m)}{(\log n)\log\log m}\right)$$

where $n$ is greater than $m$.

thanks

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I answered your clarified question below. For any $m>e^e$ there is $m_0>m$ such that $f(m,n)<0$ for any $n\in(m,m_0)$. –  GH from MO Jun 1 '13 at 18:21
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up vote 3 down vote accepted

This is my second response, after the clarifying remark from the asker (see below).

Let us assume that $m>e^e$ so that $\log\log m>0$. Then there is $m_0>m$ such that $f(m,n)<0$ for any $n\in(m,m_0)$. To see this claim, let $\epsilon>0$ be small, and consider $n$ such that $$ \log n=(\log m)^{1+\epsilon}. $$ Then $$ \frac{\log\log\log m}{\log\log\log n} \centerdot \frac{\log\log n}{\log\log m} = \frac{(1+\epsilon)\log\log\log m}{\log\log\log m+\log(1+\epsilon)} $$ $$=1+\epsilon-\frac{\epsilon}{\log\log\log m}+O(\epsilon^2)$$ while $$ \frac{\log(\frac n m)}{(\log n)\log\log m}= \frac{1-(\log m)^{-\epsilon}}{\log\log m} = \frac{\epsilon\log\log m+O(\epsilon^2)}{\log\log m}=\epsilon+O(\epsilon^2).$$ Therefore $$ f(m,n) = -\frac{\epsilon}{\log\log\log m}+O(\epsilon^2)$$ which is negative if $\epsilon>0$ is sufficiently small in terms of $m$.

P.S. My original response contained the more obvious claim that $f(m,n)>0$ when $n$ is sufficienly large in terms of $m$. Hence, for any $m>e^e$, there exist negative and positive values of $f(m,n)$ with $n>m$.

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@GH, thanks, but my question was that if there is any m such that for all n>m, f(m,n) is positive. –  asd Jun 1 '13 at 17:35
    
@asd: I have now answered your clarified question. –  GH from MO Jun 1 '13 at 18:20
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