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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. –  Ross Millikan Nov 7 '10 at 23:33
    
sorry, "fixed $n$", was imprecise, I should be able to choose e.g. $n=x$ and this bound should still hold. –  Troy K. Nov 8 '10 at 0:42

2 Answers 2

This answer corresponds to the original question, where $n$ was assumed fixed.

Considering the corresponding integral $\int_a^x {\frac{{du}}{{\log (u/n)}}} = n\int_{a/n}^{x/n} {\frac{{du}}{{\log (u)}}} $ (where $a$ is some constant) and the fact that the logaritmic integral function li(x) is $O(x/\log(x))$ as $x \to \infty$, your bound should be $O(x/\log(x))$ (and not $O(x\log(x))$.

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Thanks, sorry about the confusion with the "fixed $n$". –  Troy K. Nov 8 '10 at 1:10

Assuming $1\leq n \leq x$, split the sum into two parts, $1\leq m <n$ and $n<m\leq x$.

To handle the first part, make the variable change $m\to n-k$, and notice that $$ \sum_{1\leq m <n } \frac{1}{|\log\frac{m}{n}|} = - \sum_{1\leq k \leq n-1} \frac{1}{\log\frac{n-k}{n}}. $$ Since $-\log\frac{n-k}{n} > \frac{k}{n}$ for $1\leq k \leq n-1$, this last sum is $$ < n \sum_{1\leq k \leq n-1} \frac{1}{k} = O(n\log n) = O(x\log x).$$

The second part can be handled similarly making the variable change $m\to n+k$.

Remark: Out of curiosity, are you computing moments of the Riemann zeta-function?

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Thanks very much! At the time of writing the question, I was working through this paper, so your suspicion was correct. –  Troy K. Mar 6 '11 at 18:43

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