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I would like to know the sign of the following term in general. I tried approximately and it was negative. 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 $m$.


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up vote 2 down vote accepted

When $n$ is much larger than $m$, the function is positive because $\dfrac{\log \log n}{\log \log \log n}$ grows indefinitely as $n\to\infty$ while the term in parentheses remains bounded. But when $n$ is close to $m$ (say, $n=m+1$), the function is negative.

Indeed, $f(m,m)=0$ and the partial derivative with respect to $n$ is negative when $n=m$. I checked this with Sage:

m,n=var('m n')
f(m,n)=log(log(log(m)))/log(log(log(n)))*log(log(n))/log(log(m)) - 1 - log(n/m)/(log(n)*log(log(m))) 

The result was $$\frac{\partial f}{\partial n}(m,m) = -\frac{1}{m\log m \log \log m \log \log \log m}$$ which is obviously negative. This implies $f(m,n)<0$ for $m<n<m+\epsilon$ with some $\epsilon>0$.

Since you probably meant $n$ to be an integer, I checked to make sure $\epsilon$ will not be too small. This is done by looking at the second derivative:


I won't give the long output: it suffices to say that it has $n^2$ in the denominator. Since the first derivative is roughly $-1/m$ when $n=m$, it follows that $$\frac{\partial f}{\partial n}(m,n)<0$$ for $m<n<m+\epsilon $ where $\epsilon$ is not small: it's of order $m$ itself. Hence, $f(m,n)<0$ holds in this range too.

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