MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $m$.


share|cite|improve this question
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
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 $n>m$.

share|cite|improve this answer
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.