It is known to be false that $\sum_{m\leq x} \mu(m) \leq \sqrt{x}$ for all $x$ (Mertens' conjecture), and it is generally believed that $\lim \sup_{x\to \infty} |M(x)|/\sqrt{x} = \infty$. From the latter, it would follow that $$\lim \sup_{x\to\infty} \sqrt{x} \left|\sum_{m\leq x} \frac{\mu(m)}{m}\right| = \infty,$$ by partial summation. However, what about a smoothed sum, such as $$\sum_{m\leq x} \frac{\mu(m)}{m} \log \frac{x}{m}?$$ Is it clear that $$\lim \sup_{x\to\infty} \sqrt{x} \left|\sum_{m\leq x} \frac{\mu(m)}{m} \log \frac{x}{m} - 1\right| = \infty?$$ If one can't deduce the truth of this easily from $\lim \sup_{x\to \infty} |M(x)|/\sqrt{x} = \infty$, I'd be interested in what standard random models imply on the matter.

  • $\begingroup$ Is the sum over $m$ or $d$? $\endgroup$ – Fan Zheng Jan 25 '17 at 19:19
  • $\begingroup$ typo corrected. $\endgroup$ – H A Helfgott Jan 25 '17 at 19:19
  • $\begingroup$ Incidentally, up to where does previous work on numerics for $\sum_{m\leq x} (\mu(m)/m) \log x/m$ go? I am only aware of some calculations for small $x$ by Ramare. I'm getting unpleasant computational behavior at about $x\sim 10^{10}$. The sum is probably doing nothing actually strange; errors just start to accumulate rapidly. $\endgroup$ – H A Helfgott Jan 26 '17 at 1:41

We have that \[\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} = \frac{1}{2\pi i} \int_{\sigma_0 - i\infty}^{\sigma_0 + i\infty} \frac{1}{\zeta(s + 1)} \frac{x^s}{s^2} \, ds\] for $\sigma_0$ sufficiently large; see the bit about Riesz typical means in Section 5.1 of Montgomery and Vaughan.

Now move the contour to the left (I'm ignoring the issue of the horizontal contours - these can be dealt with, though it takes a little effort). We pick up a pole at $s = 0$ with residue $1$. From here, the standard methods (Section 15.1 in Montgomery and Vaughan) show that \[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| > 0.\]

EDIT: Lucia is right that the Riemann hypothesis together with Linear Independence hypothesis do not imply that \[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| = \infty.\] Rather, they imply that \[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| = \sum_{\rho} \frac{1}{|\zeta'(\rho)| |\rho - 1|^2}.\] To see this, assume the Riemann hypothesis and the simplicity of the zeroes of $\zeta(s)$, and mimic the proof of Lemma 4 of Ng's paper on the summatory function of the Möbius function to get an explicit expression more or less of the form \[\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1 = \sum_{\rho} \frac{x^{\rho - 1}}{\zeta'(\rho) (\rho - 1)^2} + o\left(\frac{1}{\sqrt{x}}\right),\] where the sum is over the nontrivial zeroes of $\zeta(s)$. This gives the bound \[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| \leq \sum_{\rho} \frac{1}{|\zeta'(\rho)| |\rho - 1|^2}.\] If one additionally assumes the Linear Independence hypothesis, then the Kronecker-Weyl equidistribution theorem implies that this is in fact an equality; this method goes back to the work of Ingham.

A conjecture due to Gonek and Hejhal states that \[J_{-1/2}(T) := \sum_{0 < \gamma < T} \frac{1}{|\zeta'(\rho)|} \asymp T (\log T)^{1/4}.\] Using methods from random matrix theory, Hughes, Keating, and O'Connell conjecture more generally that $J_k(T) := \sum_{0 < \gamma < T} |\zeta'(\rho)|^{2k} \sim c_k T (\log T)^{(k + 1)^2}$ whenever $\Re(k) > -3/2$, where \[c_k = \frac{1}{2\pi} \frac{G(k + 2)^2}{G(2k + 3)} \prod_p \left(1 - \frac{1}{p}\right)^{k^2} \sum_{m = 0}^{\infty} \left(\frac{\Gamma(m + k)}{m! \Gamma(k)}\right)^2 \frac{1}{p^m},\] with $G(z)$ being the Barnes $G$-function. So by partial summation, the Gonek-Hejhal conjecture implies that \[\sum_{\rho} \frac{1}{|\zeta'(\rho)| |\rho - 1|^2}\] converges.

I'm not surprised that your numerical calculations show that this is very small; Kotnik and van de Lune do numerical calculations for a similar sum over zeroes (see Table 5 of their paper) and obtain a sequence of partial sums that seem to converge to something very small.

On the other hand, if the Riemann hypothesis is false or if $\zeta(s)$ has a zero of order greater than $1$, then \[\limsup_{x \to \infty} \sqrt{x} \left|\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} - 1\right| = \infty.\] (These can both be proved using the methods in Section 15.1 of Montgomery and Vaughan.)

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    $\begingroup$ I can flesh this ought in more detail later if you'd like, but I have to ride over to Fine to meet with Sarnak at afternoon tea. $\endgroup$ – Peter Humphries Jan 25 '17 at 19:54
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    $\begingroup$ Please give him our regards! $\endgroup$ – GH from MO Jan 25 '17 at 20:36
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    $\begingroup$ I'm not sure I believe the last statement. I expect that $\sum_{\rho} 1/|\zeta^{\prime}(\rho) \rho^2|$ converges. There is a conjecture of Gonek on the size of $\sum_{0 \le \gamma \le T} 1/|\zeta^{\prime}(\rho)|$ (essentially growing like $T (\log T)^A$ for some constant $A$) from which the convergence of the other series follows. $\endgroup$ – Lucia Jan 25 '17 at 20:46
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    $\begingroup$ @HAHelfgott, GH from MO, Lucia: Sarnak says hi to the three of you. (He also asked me "Isn't MathOverflow a bit like cheating, getting other people to do your work for you? Also then everyone knows what you're working on.") $\endgroup$ – Peter Humphries Jan 25 '17 at 22:29
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    $\begingroup$ I think MO is not different from a conversation in Fine Hall :-) Well, it can be a bit addictive, but this is true of any good thing. On the other hand, one can give and get credit for assistance at MO. $\endgroup$ – GH from MO Jan 25 '17 at 23:04

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