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

Question

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 $-1/\zeta(1/2)$.

Answer 1

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.

Answer 2

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.