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Question: Is $$O\left(\frac{n}{\log(n/m)}\right)=O\left(\frac{n}{\log n}\right)$$ when $1 \leq m=o(n)$?

It would be true when $m \leq k n^\varepsilon$ for any fixed $\varepsilon < 1$ and constant $k$, since then we have $$\log (n/m) \geq \log(n/(k n^\varepsilon)) = \log (n^{1-\varepsilon}/k)= (1-\varepsilon)\log n - \log k$$ so $$\frac{n}{\log (n/m)} \leq \frac{n}{(1-\varepsilon)\log n-\log k}=O\left(\frac{n}{\log n}\right).$$ But it doesn't seem immediate how to do the case in the question.

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    Is $m$ fixed with respect to $n$? – 2017-01-18
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    Counterexample: $$m=\frac{n}{\log n}\ll n$$ then $$\frac{n}{\log(n/m)}=\frac{n}{\log\log n}\gg\frac{n}{\log n}$$ The desired property holds **if and only if** there exists some $\epsilon<1$ such that $$m=O(n^\epsilon)$$ – 2017-01-18
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    @Did: Thanks; that's fine as an answer. (And, yeah, I should have thought of that.) – 2017-01-18

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