Let $\mu(n)$ denote the Mobius function with the well-known Dirichlet series representation $$ \frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}}. $$ Basic theorems about Dirichlet series imply that if the Dirichlet series on the right converges for some $s = \sigma + it$, then it converges for all $s$ with real part $> \sigma$. Hence, $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ converging is a sufficient condition for the Riemann hypothesis.

One way to approach this question is to use partial summation. Let $M(x) = \sum_{n \leq x} \mu(n)$. Then $$ \sum_{n \leq x} \frac{\mu(n)}{\sqrt{n}} = \frac{M(x)}{\sqrt{x}} + \frac{1}{2} \int_{1}^{x} \frac{M(t)}{t^{3/2}} \, dt. $$ Odlyzko and te Riele proved that $\liminf_{x \to \infty} \frac{M(x)}{\sqrt{x}} < -1.009$ and $\limsup_{x \to \infty} \frac{M(x)}{\sqrt{x}} > 1.06$. Much earlier, Ingham had showed that $M(x)/\sqrt{x}$ was unbounded assuming the linear independence of the imaginary parts of the zeroes of $\zeta(s)$.

In addition, Gonek has an unpublished conjecture (mentioned in Ng's paper "The distribution of the summatory function of the Mobius function") that $$ -\infty < \liminf_{x \to \infty} \frac{M(x)}{\sqrt{x} (\log \log \log x)^{5/4}} < 0 <\limsup_{x \to \infty} \frac{M(x)}{\sqrt{x} (\log \log \log x)^{5/4}} < \infty. $$

Using these results and conjectures to address the original question seems to be challenging, because of possible cancellation between $\frac{M(x)}{\sqrt{x}}$ and $\int_{1}^{x} \frac{M(t)}{t^{3/2}} \, dt$. My questions are the following:

  1. Are known results about $M(x)$ enough to determine if $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges?

  2. If not, does Gonek's conjecture (or any other plausible conjectures) imply that $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges?


1 Answer 1


One can show that $\sum_{n=1}^{\infty} \mu(n)/\sqrt{n}$ diverges. Suppose to the contrary that it converges, which as you note implies RH. Put $M_0(x)=\sum_{n\le x} \mu(n)/\sqrt{n}$, and our assumption is that $M_0(x)=C+o(1)$ as $x\to \infty$.

Note that for any $s=\sigma+it$ with $\sigma>1/2$ we have $$ \int_0^{\infty} sM_0(e^x)e^{-sx} dx = \sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}} \int_{\log n}^{\infty} se^{-sx} dx = \frac{1}{\zeta(s+1/2)}. \tag{1} $$ Since $1/\zeta(s+1/2)$ is analytic (by RH) in $\sigma >0$, the identity above also holds in this larger domain. But from our hypothesis we note that the LHS above is $$ \int_0^{\infty} s(C+o(1)) e^{-sx} dx = C + o(|s|/\sigma). $$ Now take $s=\sigma+i\gamma$, where $\gamma =14.1\ldots $ is the ordinate of the first zero of $\zeta(s)$. Then the RHS of (1) is $\sim C_0/\sigma$ for a constant $C_0 \neq 0$ (essentially $1/\zeta^{\prime}(1/2+i\gamma)$). Letting $\sigma \to 0$ from above, we get a contradiction.

Note that the same heuristics underlying Gonek's conjecture should also suggest that $M_0(x)$ grows like $(\log \log \log x)^{5/4}$. I'm sure all this is classical, but I don't know a reference offhand.

  • 4
    $\begingroup$ This proof seems to work for any Dirichlet series whose analytic continuation has a pole, that is: Ordinary summation of a Dirichlet series must fail on the vertical line containing the pole. $\endgroup$ Oct 3, 2014 at 14:18
  • 4
    $\begingroup$ Lucia, let me ask a silly question : assuming the RH, on what depends the growth rate of $M_0(x)$ the more ? the asymptotic density of zeros ? their order ? the modulus of the residues ? $\endgroup$
    – reuns
    Aug 3, 2016 at 15:45

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