Let $\mu(n)$ the Möbius function. I know that combining Abel summation formula, the Prime Number Theorem and l'Hôpital's rule I can deduce $$\lim_{x\to\infty}\frac{1}{x}\sum_{2\leq n\leq x}\mu(n)\cdot\frac{\log\log n}{{\log\log (n+1)}}= 0-\lim_{x\to\infty}\frac{1}{x}\int_2^x M(t)\cdot f'(t)dt,$$ where $M(x)$ is the Mertens function, and $f(x)=\frac{\log\log x}{{\log\log (x+1)}}$. I know that $M(t)$ can be bounded by the obvious $t$, thus RHS should be

$$\lim_{x\to\infty}\frac{1}{x}O\left(\int_2^x \frac{(t+1)\log(t+1)\cdot\log\log(t+1)-t\log t\cdot\log\log t)}{(t+1)\log t\cdot\log(t+1)\cdot\log^2\log(t+1)}dt\right).$$

Question. Is there a reasonable way to compute (or compute improving my bound for Mertens function without additional more tedious computations) the limit in RHS? I say, if it is possible, determine if there is convergence and, if there is a limit compute it without do the more hardest computations. Thanks in advance.

Now with the BOUNTY I am looking an approach that provide us the behaviour of the first series, thus RHS, thus if it neccesary tedious computations your answer will be welcome. Thanks.

  • $\begingroup$ My goal with this exercise is to know with more or less precission how work with a situation of this kind, and if it is possible impove the bound for Mertens function that I've used, but it isn't neccesary give the more best result. The more important is that I want to know how work easily, if it is possible, with complicated functions doing easy computations. Thanks $\endgroup$ – user243301 Feb 17 '16 at 18:24
  • $\begingroup$ I ask: It is possible to get a bound or an asymptotic quickly, without the more tedious computations, and compute the asymptotic behaviour of RHS correctly? If there is a limit od RHS, **comopute it in the sense that say if at least is finite or $\infty$, since I know that we are work with the bound encoded in the big oh notation. $\endgroup$ – user243301 Feb 18 '16 at 6:52

Note that holds $$\frac{\log\left(\log\left(n\right)\right)}{\log\left(\log\left(n+1\right)\right)}\sim1$$ as $n\rightarrow\infty$ so your sum is very close to $M\left(x\right)$. Then $$\frac{1}{x}\sum_{2\leq n\leq x}\mu\left(n\right)\frac{\log\left(\log\left(n\right)\right)}{\log\left(\log\left(n+1\right)\right)}\sim\frac{1}{x}\sum_{2\leq n\leq x}\mu\left(n\right)=\frac{1}{x}\left(M\left(x\right)-1\right)\rightarrow0$$ as $x\rightarrow\infty$ since PNT is equivalent to $$\frac{M\left(x\right)}{x}\stackrel{x\rightarrow\infty}{\longrightarrow}0.$$ Maybe it's interesting to claim a related result which can be found in the Terence Tao blog (with the references).

Proposition: Let $f:\mathbb{N}\rightarrow\mathbb{C} $ be a bounded function such that $$\sum_{n\leq x}f\left(pn\right)\overline{f\left(qn\right)}=o_{p,q}\left(x\right) $$ for any distinct $p,q$ prime numbers, then $$\sum_{n\leq x}\mu\left(n\right)f\left(n\right)=o\left(x\right).$$

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    $\begingroup$ Very thanks much, you make that learn mathematics to be easy. Very thanks much for the reference. Can you add in a comment a definition for $o_{p,q}(x)$ or a toy example of use of the proposition? $\endgroup$ – user243301 Mar 18 '16 at 16:35
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    $\begingroup$ @user243301 My pleasure. With $o_{p,q}(x)$ I intend that the decay rate of the error term may depend on $p$ and $q$. An example, from the blog of Tao, is $$\sum_{n\leq x}\mu(n)e^{2\pi i\alpha n}=o(x)$$ where $\alpha$ is irrational. I forgot to explain that $p,q$ are prime numbers. $\endgroup$ – Marco Cantarini Mar 18 '16 at 16:40
  • $\begingroup$ Very thanks much. $\endgroup$ – user243301 Mar 18 '16 at 16:42

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