# Is the asymptotic for $\sum_n (\mu(n)/\sigma(n)) \log(x/n)$ also an upper bound?

For $x\geq 1$,

$$\rho(x) = \sum_{n\leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n}.$$

As $x\to\infty$, this sum tends to $\zeta(2) = \pi^2/6$. Is it in fact the case that $\rho(x)\leq \zeta(2)$ for all $x\geq 1$? (Computations confirm this for $x\leq 65000000$; this takes just a minute or two, but I'm using interval arithmetic, and precision issues preclude me from going further.) How would one go about proving such an inequality?

(We can derive an asymptotic expression for $\rho(x)$ from an asymptotic expression for $\check{m}(x) = \sum_{n\leq x} (\mu(n)/n) \log x/n$ (bounded by Balazard and Ramaré); the hard part is showing that the error term is negative.)

• What is the asymptotic expression? – Peter Humphries Apr 4 '17 at 19:44
• Well, at the coarsest level: $\check{m}(x)$ tends to $1$, and it is not hard to show from this that $\rho(x)$ tends to $\zeta(2)$. If GRH holds, then the error term should be $O(1/sqrt(x))$ in either case. Confusingly, it seems smaller in practice. – H A Helfgott Apr 4 '17 at 19:50
• Can you write up the details on the asymptotic expression for $\rho(x)$? – Peter Humphries Apr 4 '17 at 19:52
• Also it's not surprising that the error term for $\check{m}(x)$ is often much smaller than $O(x^{-1/2})$; it's the same reason that $|M(x)|/\sqrt{x}$ is usually smaller than $1$, namely that you have an expansion of the normalised error term of these summatory functions as (essentially) almost periodic functions of the form $\sum_{\gamma} c_{\gamma} x^{i\gamma}$, and "most" of the time, there will be a ton of cancellation in these sums. You can make this precise via the Rubinstein-Sarnak method of determining the limiting logarithmic distributions of these normalised error terms. – Peter Humphries Apr 4 '17 at 19:56
• For those of us slow on the uptake , $\mu(n)$ is the Moebius function and $\sigma(n)$ is sum of divisors? Gerhard "Making Sure Of These Things" Paseman, 2017.04.04. – Gerhard Paseman Apr 4 '17 at 20:02

We first observe that as $\frac{\mu(p^k)}{\sigma(p^k)}$ is equal to $1$ if $k = 0$, $-\frac{1}{p + 1}$ if $k = 1$, and $0$ otherwise, its Dirichlet series is $\sum_{n = 1}^{\infty} \frac{\mu(n)}{\sigma(n) n^s} = \prod_p \left(1 - \frac{1}{(p + 1) p^s}\right) = \frac{R(s)}{\zeta(s + 1)},$ where $R(s) = \prod_p \left(1 + \frac{1}{(p + 1)(p^{s + 1} - 1)}\right).$ Note that for $\sigma > -1$, the terms in the Euler product for $R(\sigma)$ are of the form $1 + a_p(\sigma)$ with $a_p(\sigma) = \frac{1}{(p + 1)(p^{\sigma + 1} - 1)}$. Since $\sum_p a_p(\sigma)$ converges for $\sigma > -1$, it follows that $R(s)$ is absolutely convergent for $\Re(s) > -1$ and defines a holomorphic function in that region. Moreover, $R(0) = \zeta(2) = \frac{\pi^2}{6}$.

Let $\Theta$ denote the supremum of the real parts of the zeroes of $\zeta(s)$. Suppose in order to obtain a contradiction that $\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} < x^{-1 + \Theta - \varepsilon}$ for all $x > x_{\varepsilon}$. Then Landau's lemma implies that if $\sigma_c$ is the infimum of $\sigma \in \mathbb{R}$ for which $\int_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} + \frac{\pi^2}{6}\right) x^{-\sigma} \, \frac{dx}{x}$ is convergent, then $\int_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} + \frac{\pi^2}{6}\right) x^{-s} \, \frac{dx}{x}$ is holomorphic in the right half-plane $\Re(s) > \sigma_c$ but not at the point $\sigma_c \in \mathbb{R}$. On the other hand, this integral is equal to $\frac{1}{s + 1 - \Theta + \varepsilon} - \frac{R(s)}{s^2 \zeta(s + 1)} + \frac{\zeta(2)}{s}$ for $\Re(s) > 0$ and hence for $\Re(s) > \sigma_c$ by analytic continuation. However, this expression has a pole at $s = -1 + \Theta - \varepsilon$ and no other poles on the real line segment $\sigma > -1 + \Theta - \varepsilon$, yet by the definition of $\Theta$, there are poles in the strip $-1 + \Theta - \varepsilon < \Re(s) \leq -1 + \Theta$. Thus a contradiction is obtained, and so it follows that $\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} = \Omega_+\left(x^{-1 + \Theta - \varepsilon}\right).$ The same method shows that $\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} = \Omega_-\left(x^{-1 + \Theta - \varepsilon}\right),$ so that $\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6}$ changes sign infinitely often. With a little extra work, we can show that $\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} = \Omega_{\pm}\left(x^{-1/2}\right).$

All of this raises the question:

Why does there seem to be a bias?

In other problems like disproving Pólya's conjecture, the bias stems from a term of the same order as the oscillatory terms coming from the zeroes of $\zeta(s)$. But this is not the case in this setting; as Greg mentions in the comments, if we additionally assume the Riemann hypothesis and the Linear Independence hypothesis, then we can show (modulo some details about the growth of $R(s)$ on the line $\Re(s) = -1/2$) that the set $\left\{x \in [1,\infty) \colon \sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6} < 0\right\}$ has a limiting logarithmic density that is equal to $1/2$.

So where does the bias come from? I believe it's due to a lower order term. As $s$ tends to $-1$, $1/\zeta(s+1)$ tends to $-2$, whereas $R(s)$ is positive and blows up as $s \to -1$. In fact, it blows up in two ways: one way from the prime $p = 2$, and the other from the product over the remaining primes. This suggests that $R(s) \sim C/(s + 1)^2$ as $s \to -1$ for some positive constant $C > 0$. So this will give a lower order term of size $-\frac{C' \log x}{x}$ in the asymptotic expansion of $\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n} - \frac{\pi^2}{6}$, where $C' > 0$ is some positive constant.

• Very nice. If you assume RH and the linear independence of $\{\gamma>0\colon \zeta(\frac12+i\gamma)=0\}$ over $\Bbb Q$, can you prove that the logarithmic density of those $x$ for which the sum exceeds $\frac{\pi^2}6$ is actually $\frac12$? Or is there a bias-causing term somewhere in the explicit formula? (I don't see one....) – Greg Martin Apr 5 '17 at 8:51
• Good call. A more precise computation shows that the proposed inequality is false for x = 65371730 (and possibly for some other x> 65371719; for x<=65371719, it is true) – H A Helfgott Apr 5 '17 at 15:13
• 30, or 20? Gerhard "Or Nineteen Or Twenty Nine?" Paseman, 2017.04.05. – Gerhard Paseman Apr 5 '17 at 15:44
• 30. The gap should be easy to remove by the use of higher-precision arithmetic. The calculation I'm doing uses an interval-arithmetic package, and thus is rigorous, but the package is based on merely double-precision arithmetic (for speed). Hence the (small) gap, even after some careful optimizations. – H A Helfgott Apr 5 '17 at 21:24
• It's interesting to see what happens if we impose the condition that $n$ run over the odd numbers only. The limit is then $\pi^2/4$. It is an upper bound for much longer: it is valid for $x<=1465984277$ (and possibly a bit further). – H A Helfgott Apr 6 '17 at 5:40