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I am trying to reproduce the following bound:

$\sum_{1\leq m\leq x, m\neq n}\frac{1}{|\log(m/n)|}=O(x\log(x))$,

for $x\geq 2$ and some $n$, $1\leq n\leq x$ (the implicit constant shouldn't depend on $n$).

I have tried to bound each term by $\frac{1}{|\log(1\pm 1/x)|}$, but this doesn't seem to be good enough. Splitting the sum at $\log(x)$ also didn't help. Any ideas?

Thank you very much for your comments or hints.

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    I think you may need to split it at kn for some constant k. The sum up to kn will not depend on x, so is negligible as x gets very large (but will grow with n). Above kn you can change over to an integral. Hope this helps-I haven't checked that it works.2010-11-07
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    sorry, "fixed $n$", was imprecise, I should be able to choose e.g. $n=x$ and this bound should still hold.2010-11-08

2 Answers 2