# Properties of Mertens Function

I am astounded by how little information about Mertens function M(n) (partial sums of the Möbius function) is on the Internet. Thus, I would be thankful if someone could clear up some of my confusion.

First, I learned that PNT (prime number theorem) $\iff M(n)/n \rightarrow 0$ as $n\rightarrow \infty$

This makes sense as M(n) is the count of square-free integers up to n that have an even number of prime factors, minus the count of those that have an odd number, and I would expect these to cancel out in their contribution to the quotient as $n \rightarrow \infty$.

If |M(n)| is bounded by B, couldn't we conclude $M(n) = O(B)$? If not, then is M(n) finite for all n but unbounded? I know $M(n)<n<\infty$ for all n.

Furthermore, does anyone happen to know the best big O for M(n)? Does anyone know any online sources that exposit on M(n)?

I am thankful to anyone that can provide some information.

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I think PNT is equivalent to $M(n)=o(n)$. Anyway, where the internet fails, there is always the library. There are many books where you will find answers to your questions. – Gerry Myerson Feb 27 '13 at 2:04
We can, at the very least, say M(n)<n for all n just by the very definition of M(n), without referring to PNT at all. – Just Some Old Man Feb 27 '13 at 2:15
I am sorry, you are absolutely right. My concept of "little o" was wrong. I corrected it in my question. – Just Some Old Man Feb 28 '13 at 23:41

1. $n^{1/2} \to \infty$ as $n\to \infty$, is there any problem with that?

2. You can conclude that $M(n) = O(1)$ (or $O(B)$ if you want), if your $B$ is the same for all $n$.

3. Getting information about $M(n)$ amounts to knowing zero-free region of $\zeta(s)$ by Perron's formula, so you would want to look up zero-free region results of $\zeta(s)$. Assume Riemann Hypothesis though, Soundararajan proved that $$M(n) << \sqrt{n} exp((\log n)^{1/2} (\log \log n)^{14})$$

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1) I made an error, but the contradiction still exists. In one case, M(n) approaches infinity (unless our knowledge of M(n) is so bad that our best big O goes to infinity though M(n) goes to zero). In the other case, M(n) approaches 0. 2) M(n) cannot be O(1) (i.e. O(B)) because that would imply M(n) is O(n^{1/2} + epsilon), which means, according to Wikipedia, we have just proven the Riemann Hypothesis. – Just Some Old Man Feb 27 '13 at 3:31
@David, 1. Ah, I read your PNT line wrongly. I believe that PNT is equivalent to $M(n) = o(n)$, rather than going to 0. 2. It was unclear what your $B$ is (constant? function?). So I assumed it meant a constant. – user27126 Feb 27 '13 at 4:13

Sorry for the late answer on this, but the OP asked to the best big $O$ bound available, and it seems to be unanswered. I thought it might be useful to have here for future reference.

The best unconditional big $O$ bound that I am aware of is originally due to Walfisz, but since his book is out of print, it is best to go to Ivic's book, Theorem 12.7, for the statement

There is an absolute constant $C>0$ such that $$M(x) = \sum_{n \leq x} \mu(n) \ll x \exp(-C \log^{3/5}x(\log \log x)^{-1/5}).$$

and proof.

In addition, some explicit bounds can be found here.

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I did a Google search for "Mertens" and got these:

http://mathworld.wolfram.com/MertensFunction.html

http://mathworld.wolfram.com/MertensConjecture.html

These seem like a good start.

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Yeah, I saw them. Ironically, the article on Mertens conjecture gives more information about Mertens function than the article on Mertens function. – Just Some Old Man Feb 27 '13 at 2:39