I’ve just uploaded to the arXiv the paper A remark on partial sums involving the Möbius function, submitted to Bull. Aust. Math. Soc..

The Möbius function is defined to equal when is the product of distinct primes, and equal to zero otherwise; it is closely connected to the distribution of the primes. In 1906, Landau observed that one could show using purely elementary means that the prime number theorem

(where denotes a quantity that goes to zero as ) was logically equivalent to the partial sum estimates

we give a sketch of the proof of these equivalences below the fold.

On the other hand, these three inequalities are all easy to prove if the terms are replaced by their counterparts. For instance, by observing that the binomial coefficient is bounded by on the one hand (by Pascal’s triangle or the binomial theorem), and is divisible by every prime between and on the other hand, we conclude that

from which it is not difficult to show that

Also, since , we clearly have

Finally, one can also show that

Indeed, assuming without loss of generality that is a positive integer, and summing the inversion formula over all one sees that

and the claim follows by bounding by .

In this paper I extend these observations to more general multiplicative subsemigroups of the natural numbers. More precisely, if is any set of primes (finite or infinite), I show that

where is the multiplicative semigroup generated by , i.e. the set of natural numbers whose prime factors lie in .

Actually the methods are completely elementary (the paper is just six pages long), and I can give the proof of (7) in full here. Again we may take to be a positive integer. Clearly we may assume that

as the claim is trivial otherwise.

If denotes the primes that are *not* in , then Möbius inversion gives us

Summing this for gives

We can bound and so

The claim now follows from (9), since and overlap only at .

As special cases of (7) we see that

and

for all . Since , we also have

One might hope that these inequalities (which gain a factor of over the trivial bound) might be useful when performing effective sieve theory, or effective estimates on various sums involving the primes or arithmetic functions.

This inequality (7) is so simple to state and prove that I must think that it was known to, say, Landau or Chebyshev, but I can’t find any reference to it in the literature. [*Update*, Sep 4: I have learned that similar results have been obtained in a paper by Granville and Soundararajan, and have updated the paper appropriately.] The proof of (8) is a simple variant of that used to prove (7) but I will not detail it here.

Curiously, this is one place in number theory where the elementary methods seem superior to the analytic ones; there is a zeta function associated to this problem, but it need not have a meromorphic continuation beyond the region , and it turns out to be remarkably difficult to use this function to establish the above results. (I do have a proof of this form, which I in fact found before I stumbled on the elementary proof, but it is far longer and messier.)

** — 1. Equivalences — **

We now show the equivalences between (1), (2), and (3). The arguments here are drawn from this article of Diamond; I will skip some of the details in order to focus on the main ideas.

It is convenient to introduce the von Mangoldt function , defined as when is a prime (or a power of that prime), and zero otherwise. It is not difficult to see that the prime number theorem (1) is equivalent to the asymptotic

Now we show that (10) implies (2). The starting point is the identity

connecting and , which is the coefficient form of the identity

involving the Riemann zeta function . Summing this identity for , one obtains the formula

Using the Chebyshev bound (4), one can bound by ; using (10), one can refine this to for any when is large enough. Using these bounds together with (5), one can show that the right-hand side of (11) is . It is not too difficult to get to (2) from here (e.g. by summation by parts).

To get from (2) to (3), we use (6). A summation by parts argument using (2) shows that

for any fixed (basically because the function has bounded variation). Letting go to infinity and using (6) we obtain (3). Conversely, a summation by parts argument easily deduces (2) from (3).

Finally, to get back from (2) and (3) to (10), we use the identities

where is the divisor function; these identities are the coefficient forms of the identities

and

Expanding out using these formulae, we see that to prove (10) it suffices to show that

From the Dirichlet hyperbola method one can show that

for some absolute constant (in fact , where is Euler’s constant), and from Stirling’s formula one has

and so

From this, (2), (3), and summation by parts one soon obtains (14).

Remark 1These equivalences are elementary, but somewhatad hocin nature. A less elementary, but more “natural”, approach would be to rewrite everything in terms of the Riemann zeta function, and in particular to show that (10), (2), (3) are all equivalent to the assertion that the zeta function has a simple pole at and no other poles or zeroes on the line . This approach can of course be found in any introductory analytic number theory text.

## 16 comments

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30 August, 2009 at 9:21 pm

AnonymousTerry:

Should this be tagged as ‘paper’ instead of ‘expository’?

[Corrected, thanks. -T]31 August, 2009 at 8:46 am

AnonymousDear Prof. Tao,

The spelling of Hans von Mangoldt’s name seems to be incorrect, in the second paragraph of section 1. (Feel free to delete this nitpicky post, also!)

[Corrected, thanks - T.]31 August, 2009 at 9:24 am

AnonymousDear Terry,

The arxiv link to the paper doesn’t seem to be the right one.

[Oops, the paper hasn't appeared on the arXiv yet. It should within a day or so. -T]31 August, 2009 at 11:03 am

AnonymousI don’t see how you get the number 1 on the far left of equation (6). Should that be x instead? (in which case you would get 2, not 1, as the desired bound).

31 August, 2009 at 12:43 pm

Terence TaoThe sum equals 1 for n=1 and zero otherwise, so is equal to 1 for . This sum can then be rearranged as .

31 August, 2009 at 9:28 pm

AnonymousThanks!

1 September, 2009 at 1:06 am

Andrew V. SutherlandBelow is a simple example that I believe proves your conjecture that the sum

S(G,x) = sum_{n \ in G : n <= x} mu(n)/n

diverges for certain mutliplicative subsemigroups G of the natural numbers.

Let P be the set of positive integers n=pq that are the product of two primes, and let G=. For an element n of G, we have mu(n)=1 when n is square-free and mu(n) = 0 otherwise. In particular, mu(n) >= 0 for all n \in G, hence if S(G,x) converges, it does so absolutely.

But note that S(G,x) is bounded below by 1/2 sum_{2<p<=x} 1/p, which diverges, by Mertens' second theorem (sum_{p<=x} 1/p ~ loglog(x)).

1 September, 2009 at 1:09 am

Andrew V. SutherlandHere G = \langle P \rangle is the semigroup generated by P.

1 September, 2009 at 1:32 am

Terence TaoAh, yes, that seems to work, thanks!

2 September, 2009 at 11:23 am

Kevin O'BryantIf you search for “Beurling numbers” on Google (or Math SciNet, I suppose, although I don’t have access at the moment), you’ll find many related articles. Harold Diamond at the University of Illinois is an expert.

If I recall correctly, Beurling numbers are defined in the following manner. Let {p_1<p_2 < …} be an increasing sequence of real numbers, all larger than 1. These are our "Beurling primes". The "Beurling numbers" are the semigroup generated by the p_i, counting mulitiplicity. The general idea is that if the Beurling numbers are enough like the positive integers, then the Beurling primes are very similar to the primes. For example, there's a theorem along the lines of "If the number of Beurling numbers less than x is asymptotically cx (for any positive constant c, with a nice error term), then the number of Beurling primes less than x is asymptotically x/log(x)." In other words, the prime number theorem isn't really about the integers, it's just talking about semigroups of reals with linear growth.

Interestingly, the PNT error term can provably not be improved in this setting. It seems that any progress on the error term, the Riemann Hypothesis included, *must* utilize some properties of the integers aside from their multiplicative structure.

I learned about this line from a grad student at Illinois (Italian, but I don't recall his name) whose thesis was showing that for almost all (in some sense) choices of p_i, the Riemann Hypothesis does hold for the consequent zeta function.

I just found http://www.math.kth.se/~rikardo/beurling.pdf, which seems relevant.

2 September, 2009 at 12:39 pm

AnonymousO’Bryant post is very interesting. However, in your next to last paragraph, do you mean “the Riemann Hypothesis doesn’t hold”? I would very much want to see the paper if you really mean otherwise.

6 September, 2009 at 8:40 pm

Kevin O'BryantUnfortunately, I don’t have any references. What I’ve heard has been in casual conversation and at seminars. The Italian student I referred to above is Eugene Balanzario, but I didn’t find any telling links with his name.

My impression is that the Riemann Hypothesis tends to hold (if the counting function for the Beurling numbers is ‘enough’ linear) in some rigorous sense, but that there are Beurling numbers with very linear growth where it fails. But, this is now third-hand info, that may not have been refereed to begin with. Harold Diamond is the one who should weigh in with the truth of things, what’s known true and known false.

15 September, 2009 at 1:34 pm

Eugenio P. BalanzarioKevin, my second given name is Pacelli, which is certainly Italian, but I am not. You can easily produce examples of Beurling numbers for which the Riemann Hypothesis is either true or false. In my PhD thesis, written under the direction of Prof. Diamond, I considered the Mobius function of a particular set of Beurling numbers. This example is interesting because it shows that the prime number theorem is not equivalent (in the general setting of Beurling) to the statement M(x)=o(x) where M(x) is the sum of the Moubius function.

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