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Let $\mu(k)$, the Möbius function. I've deduced some expressions for the Mertens function, that is $$M(N)=\sum_{k\leq N}\mu(k),$$ by specialization of Möbius inversion for Taylor series: since $ \left| \frac{\mu(k)}{2} \right| <1$ for each $k\geq 1$ then one can writes by Möbius inversion for Taylor series the following identity $$\frac{\mu(k)}{2}=-\sum_{n=1}^\infty\frac{\mu(n)}{n}\log \left( 1- \frac{(\mu(k))^n}{2^n}\right).$$ Thus if there are no mistakes one can deduces $$M(N)=2 \sum_{k=1}^{N} \sum_{n=1}^{\infty}\frac{\mu(n)}{n} \sum_{m=1}^{\infty}\frac{(\mu(k))^{nm}}{m2^{nm}} . $$

I am interested in what kind of calculations one can do from these identities that can be deduced from this inversion formula (for a different example that I am trying see $\dagger$), that is: I know that it is easy to deduce the obvious bound $M(N)=O(N)$ using the triangle inequality and the big oh notation. Also I know that there were in the literature better bounds like as $O(\frac{N}{1+\log^2 N})$ (also it is well known the conjecture related with Mertens function), but I will ask

Question. Can you say if $$M(N)=2 \sum_{k=1}^{N} \sum_{n=1}^{\infty}\frac{\mu(n)}{n} \sum_{m=1}^{\infty}\frac{(\mu(k))^{nm}}{m2^{nm}} . $$ was right, and deduce the asymptotic behaviour as $N$ tends to $\infty$, I say as a big oh statement, for $$2 \sum_{k=1}^{N} \sum_{n=1}^{\infty}\frac{\mu(n)}{n} \sum_{m=1}^{\infty}\frac{(\mu(k))^{nm}}{m2^{nm}} $$ without using previous identity (see my attempt)? Thanks in advance.

My attempt (I don't know if in my attempt were mistakes) was that using the triangle inequality $$2 \left|\sum_{k=1}^{N} \sum_{n=1}^{\infty}\frac{\mu(n)}{n} \sum_{m=1}^{\infty}\frac{(\mu(k))^{nm}}{m2^{nm}} \right| \leq 2 \sum_{k=1}^{N} \sum_{n=1}^{\infty}\frac{1}{n} \sum_{m=1}^{\infty}\frac{1}{m2^{nm}} ,$$ and this last RHS is equal to $$-2 \sum_{k=1}^{N} \sum_{n=1}^{\infty}\frac{1}{n}\log \left( 1-\frac{1}{2^n} \right) .$$ I am asking if it is possible to show that this is $O(N)$, because I am looking to obtain a big oh statement (an easy deduction) for such series (that is Mertens function) without use the identity with Mertens function (you can use all statements that you know, but is forbidden to use the identity that was deduced from the cited inversion formula).

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  • $\begingroup$ $\dagger$ in a different exercise I take the specizalition $\frac{1}{d}$, where $1<d\mid m$ is a divisor of a fixed integer $m$, I take the sum over all (such) divisors and after I try apply Gronwall's theorem for the sum of divisors function looking if it is possible do more calculations with $\limsup$ and the series. You are welcome if want ask me in the Chat, about this exercise for the sum of divisor function. $\endgroup$ – user243301 Aug 6 '16 at 0:24
  • $\begingroup$ As was said I am asking to obtain from this expression for the Mertens function an asymptotic likes $O(N)$ or a few better, as a deduction of such expression as an infinite series. It is feasible? Thanks. $\endgroup$ – user243301 Aug 13 '16 at 16:09
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Note that the inner sum in your last double sum doesn't depend on $k$; so either the inner sum $\sum_{n=1}^\infty \frac1n\log(1-\frac1{2^n})$ diverges, or it converges and your double sum is $O(N)$. So this isn't really about asymptotics at all, just the convergence of one sum.

In that inner sum, $\frac1n\le1$ and $0\le-\log(1-\frac1{2^n})\le \frac1{2^n}\log4$. (This latter inequality follows from the fact that the convex function $-\log(1-x)$ lies under its secant line on $[0,\frac12]$; plug in $x=\frac1{2^n}$.) Therefore the inner sum is $\le 2 \sum_{n=1}^\infty \frac1{2^n}\log4 = 2\log4$.

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  • $\begingroup$ Very thanks much, and I'm sorry by my mistake, I've seen very good mathematics in this site, so my mathematics and reasoning should be improve. $\endgroup$ – user243301 Aug 13 '16 at 20:40
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Your identity about the Mertens function is right. Using the inversion, since $$\log\left(1-x\right)=-\sum_{n\geq1}\frac{x^{n}}{n} $$ we have $$x=-\sum_{n\geq1}\frac{\mu\left(n\right)}{n}\log\left(1-x^{n}\right) $$ so if we take $x=\mu\left(k\right)/2 $ and summing from $1$ to $N$ we can conclude. About the bound $O(N)$, as noted by Greg Martin, it is sufficient to show that the series is covergent. Another approach can be following. Note that $$ S=-\sum_{n\geq1}\frac{1}{n}\log\left(1-\frac{1}{2^{n}}\right)=\sum_{n\geq1}\frac{1}{n}\sum_{m\geq1}\frac{1}{m2^{nm}}=\log\left(2\right)+\sum_{n\geq2}\frac{1}{n}\sum_{m\geq1}\frac{1}{m2^{nm}} $$ $$=\log\left(2\right)+\sum_{m\geq1}\frac{1}{m}\sum_{n\geq2}\frac{1}{n2^{nm}}=2\log\left(2\right)-\frac{1}{2}+\sum_{m\geq2}\frac{1}{m}\sum_{n\geq2}\frac{1}{n2^{nm}} $$ and since $$nm\geq n+m,\, n,m\geq2 $$ we have $$S\leq2\log\left(2\right)-\frac{1}{2}+\sum_{m\geq2}\frac{1}{m2^{m}}\sum_{n\geq2}\frac{1}{n2^{n}}\leq4\log\left(2\right)-\frac{3}{2}$$ so the series converges.

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  • $\begingroup$ Very thanks much also for your detailed answer. $\endgroup$ – user243301 Aug 13 '16 at 20:41

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