# Is it possible to prove $\lim_{ x \to \infty}M(x)/\sqrt{x} - \sum_{n \le x}\mu(n)/\sqrt{n}$ converges?

With $$\mu(n)$$ being Möbius function and $$M(x) = \sum_{n \le x} \mu(n)$$ being Martens function:

Is it possible to prove that the difference between two diverging terms given below $$\lim_{ x \to \infty} \Big ( \frac{M(x)} {\sqrt{x}} - \sum_{n \le x} \frac{\mu(n)}{\sqrt{n}} \Big)$$ converges?

Both terms above diverge but, computationally, the difference between them seems to be converging to $$-1/\zeta(1/2)$$.

• The limit does not converge. It oscillates around $-1/\zeta(1/2)$. Jul 21, 2021 at 3:05

This conjecture is false. To disprove this, we must first consider two possible cases.

First, if RH is false, then the supremum $$\Theta$$ of the imaginary parts of the zeroes of $$\zeta(s)$$ is larger than $$1/2$$. We claim that $$x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2}$$ changes sign infinitely often. Suppose in order to obtain a contradiction that there exists some $$0 < \varepsilon < \Theta - 1/2$$ and $$x_{\varepsilon} > 1$$ such that $$x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2} < x^{\Theta - 1/2 - \varepsilon}$$ for all $$x > x_{\varepsilon}$$. Then Landau's lemma (Lemma 15.1 of Montgomery-Vaughan) states that if $$\sigma_c$$ is the infimum of $$\sigma \in \mathbb{R}$$ for which $$\int_{1}^{\infty} \left(x^{\Theta - 1/2 - \varepsilon} - x^{-1/2} M(x) + \sum_{n \leq x} \mu(n) n^{-1/2}\right) x^{-\sigma} \, \frac{dx}{x}$$ is convergent, then $$\int_{1}^{\infty} \left(x^{\Theta - 1/2 - \varepsilon} - x^{-1/2} M(x) + \sum_{n \leq x} \mu(n) n^{-1/2}\right) x^{-s} \, \frac{dx}{x}$$ is holomorphic in the right-half plane $$\Re(s) > \sigma_c$$, but not at the point $$s = \sigma_c$$. On the other hand, this integral is equal to $$\frac{1}{s + 1/2 - \Theta + \varepsilon} + \frac{1}{2s(s + 1/2)\zeta(s + 1/2)}$$ for $$\Re(s) > 1/2$$ and hence $$\Re(s) > \sigma_c$$ by analytic continuation. However, this expression has a pole at $$s = \Theta - 1/2 - \varepsilon$$ and no other poles on the real line segment $$\sigma > \Theta - 1/2 - \varepsilon$$, yet by the definition of $$\Theta$$, there are poles in the strip $$\Theta - 1/2 - \varepsilon < \Re(s) \leq \Theta - 1/2$$. Thus a contradiction is obtained. It follows that $$\limsup_{x \to \infty} \frac{x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2}}{x^{\Theta - 1/2 - \varepsilon}} \geq 1.$$ The exact same argument shows that the limit inferior is at most $$-1$$, which means that the limit as $$x \to \infty$$ of $$x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2}$$ does not exist.

Now we suppose that RH is true. Let $$\rho = 1/2 + i\gamma$$ be a simple zero on the line $$\Re(s) = 1/2$$. Suppose in order to obtain a contradiction that there exists some $$C < |\gamma \rho \zeta'(\rho)|^{-1}$$ and $$x_0 > 1$$ such that $$x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2} + 1/\zeta(1/2) < C$$ for all $$x > x_0$$. Then for $$\Re(s) > 1/2$$, $$\int_{1}^{\infty} \left(C - x^{-1/2} M(x) + \sum_{n \leq x} \mu(n) n^{-1/2} - 1/\zeta(1/2)\right) x^{-s} \, \frac{dx}{x} = F(s),$$ where $$F(s) = \frac{C}{s} + \frac{1}{2s(s + 1/2)\zeta(s + 1/2)} - \frac{1}{s\zeta(1/2)}.$$ Then $$\int_{1}^{\infty} \left(C - x^{-1/2} M(x) + \sum_{n \leq x} \mu(n) n^{-1/2} - 1/\zeta(1/2)\right) \left(1 + \cos(\phi - \gamma \log x)\right) x^{-s} \, \frac{dx}{x}$$ is equal to $$F(s) + \frac{1}{2} \left(e^{i\phi} F(s + i\gamma) + e^{-i\phi} F(s - i\gamma)\right),$$ where $$\phi = \pi - \arg\left(\frac{1}{i\gamma \rho \zeta'(\rho)}\right).$$ We consider the limit as $$s$$ approaches $$0$$ from the right. After multiplying by $$s$$, the expression in terms of $$F(s)$$ converges to $$C - \left|\frac{1}{\gamma \rho \zeta'(\rho)}\right|.$$ By assumption, this is negative. On the other hand, this must mean that the integral above tends to negative infinity as $$s$$ approaches $$0$$ from the right. This, however, cannot be the case, as by assumption the integrand is nonnegative for all sufficiently large $$x$$. Thus a contradiction is obtained, and so $$\limsup_{x \to \infty} \left(x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2} + 1/\zeta(1/2)\right) \geq \left|\frac{1}{\gamma \rho \zeta'(\rho)}\right|.$$ The same argument shows that $$\liminf_{x \to \infty} \left(x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2} + 1/\zeta(1/2)\right) \leq -\left|\frac{1}{\gamma \rho \zeta'(\rho)}\right|.$$ This means that the limit as $$x \to \infty$$ of $$x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2}$$ does not exist.

I should note that if one assumes some standard conjectures, $$x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2}$$ does remain bounded as $$x \to \infty$$, and the limit inferior and limit superior can be explicitly written down, namely $$\limsup_{x \to \infty} \left(x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2}\right) = -\frac{1}{\zeta(1/2)} + \sum_{\rho} \left|\frac{1}{\gamma \rho \zeta'(\rho)}\right|$$ and $$\liminf_{x \to \infty} \left(x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2}\right) = -\frac{1}{\zeta(1/2)} - \sum_{\rho} \left|\frac{1}{\gamma \rho \zeta'(\rho)}\right|,$$ and this sum over nontrivial zeroes $$\rho$$ is finite.

• Thanks for the detail answer. I am in process of digesting this, but to start with, I am confused with statement “If RH is false ... we claim that difference in terms changes sign infinitely often”. Are you saying it follows from RH being false? Or are you supposing the sign will change infinitely often for s = 1/2? Jul 21, 2021 at 16:59
• If RH is false, then it changes sign infinitely often. If RH is true, then it also changes sign infinitely often. So yes, I’m claiming that it follows from RH being false, and it also follows from RH being true. Since one of these two must happen, it follows unconditionally. Jul 21, 2021 at 17:24
• How do we claim it changes signs often? May be it stays positive always for some $x \gt x_0$. I found in my computations that the difference is Always positive for $9 < x < 10^9$, and it seems to stay within 1% of 0.684765236. So it very likely may not change sign. The entity obtained by adding $1/\zeta(1/2)$ from the difference will likely change sign infinitely often. Did you mean to say this? Jul 21, 2021 at 19:46
• I proved that $x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2} + 1/\zeta(1/2)$ changes sign infinitely often and that this does not have a limit as $x \to \infty$. However I would conjecture that $x^{-1/2} M(x) - \sum_{n \leq x} \mu(n) n^{-1/2}$ is always positive for all sufficiently large $x$. Proving this is way out of reach, however. Jul 21, 2021 at 19:49
• @SungjinKim the standard conjectures are the Riemann hypothesis, simplicity of the zeroes, the linear independence hypothesis, and the Gonek-Hejhal conjecture $J_{-1}(T) := \sum_{0 < \gamma \leq T} |\zeta'(\rho)|^{-2} \ll T$. Jul 22, 2021 at 2:13

We can apply the method by Lucia at MO: https://mathoverflow.net/questions/164874/is-it-possible-to-show-that-sum-n-1-infty-frac-mun-sqrtn-diverg?noredirect=1&lq=1

to prove the divergence in a quick argument. The result does not provide the sign changes in Peter Humphries' answer. So, this is a weaker result in a quicker argument.

Let $$M(x)=\sum_{n\leq x} \mu(n)$$ and $$M_0(x)=\sum_{n\leq x} \frac{\mu(n)}{\sqrt n}$$.

We have $$\frac{-1/2}{(s+1/2)s\zeta(s+1/2)}=\int_1^{\infty} \frac{ \frac{M(t)}{\sqrt t}-M_0(t)}{t^{s+1}} dt, \ \ \ \sigma>1/2. \ \ \ (*)$$ Suppose that $$\frac{M(t)}{\sqrt t}-M_0(t)= c + o(1)$$ (the convergence the limit of OP). This implies RH. Then by RH, we have (*) for $$\sigma>0$$. Let $$\rho = 1/2 + i t_0$$ be the first nontrivial zeta zero ($$t_0 \approx 14.134725141734695$$).

We fix $$t=t_0$$ and let $$\sigma \rightarrow 0+$$. Then we have $${\rm LHS}\sim \frac{-1/2}{(1/2+it_0)it_0\sigma \zeta'(\rho)}$$ However, $${\rm RHS}=\frac c{\sigma+it_0} + o(1/\sigma).$$ This is a contradiction. Hence, we must have the divergence of the limit of OP.

• That's a very nice argument! Jul 21, 2021 at 21:37