8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
9
$\begingroup$

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.)

$\endgroup$
  • 3
    $\begingroup$ 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
  • 3
    $\begingroup$ Please give him our regards! $\endgroup$ – GH from MO Jan 25 '17 at 20:36
  • 5
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 4
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.