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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 but unbounded? It can never be infinite because $M(n)<n<\infty$ for all n.

Furthermore, does anyone happen to know the best big O M(n) is? Does anyone know any online sources that exposits 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. –  David 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. –  David Feb 28 '13 at 23:41
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2 Answers

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. –  David Feb 27 '13 at 2:39
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  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. –  David 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. –  Sanchez Feb 27 '13 at 4:13
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