2
$\begingroup$

It is mentioned in multiple occasions here that the bound $$ \mathop{\sum_{n=1}^{N}}_{n\equiv a\mod l} \mu(n) = o(N) $$ is equivalent to the prime number theorem in arithmetic progressions. But I am not able to find a proof of this equivalence (or the proof of the above bound) except for the case when $l=1$ which is the classical prime number theorem. If anyone can point me to a reference, it will be much appreciated.

Also, there is a result in Edwards' book where he proves the result $$ \left|\sum_{n=1}^{N} \frac{\mu(n)}{n}\right| \leq \frac{K}{\log(N)}. $$ Is there an analogue of this bound for arithmetic progressions as well?

There are some discussions here, but they don't consider the Möbius function

$\endgroup$
3
  • 2
    $\begingroup$ It shouldn't be equivalent to the infinitude, but rather be equivalent to the asymptotic. $\endgroup$
    – Will Sawin
    Jul 17 at 14:22
  • $\begingroup$ @WillSawin Yes, that seems more plausible. In Tao’s post it just mentions something along the lines of “primes number theorem in arithmetic progressions” Now that I think about it, it does seem that it should be equivalent to the asymptotic. I will edit my question. $\endgroup$ Jul 17 at 16:20
  • 1
    $\begingroup$ Concerning the link in the last sentence of your post, I have now added a remark about the Moebius fucntion and a link to this page at the end of my answer there. $\endgroup$
    – KConrad
    Jul 18 at 1:51

1 Answer 1

5
$\begingroup$

When $(a,m) = 1$, let $\pi(x;a \bmod m)$ be the number of primes $p\leq x$ such that $p \equiv a \bmod m$. The prime number theorem for arithmetic progressions mod $m$ says for all $a \in (\mathbf Z/m\mathbf Z)^\times$ that $\pi(x;a \bmod m) \sim (1/\varphi(m))x/\log x$.

Harold Shapiro, in the paper Some assertions equivalent to the prime number theorem for arithmetic progressions, Comm. Pure Appl. Math. 2 (1949), 293-308, showed that theorem for an integer $m \geq 1$ is equivalent to each of the following conditions for the same $m$: $$ \sum_{\substack{n \leq x \\ n \equiv a \bmod m}} \mu(n) = o(x) $$ for all $a$ with $(a,m) = 1$ and $$ \sum_{\substack{n \geq 1 \\ n \equiv a \bmod m}} \frac{\mu(n)}{n} \ \ {\rm converges} $$ for all $a$ with $(a,m) = 1$. This should address both of your questions.

Concerning the second equivalent condition above, note that the prime number theorem is in many places expressed as being equivalent to the calculation $\sum \mu(n)/n = 0$, but it is also equivalent just to the convergence of $\sum \mu(n)/n$ because it is easy to show that this series must be $0$ if it converges thanks to Abel's theorem for Dirichlet series: convergence of $\sum \mu(n)/n$ implies this series must equal $\lim_{s \to 1^+} \sum \mu(n)/n^s = \lim_{s \to 1^+} 1/\zeta(s) = 0$.

I explained in my answer here how the prime number theorem in arithmetic progressions mod $m$ is equivalent to nonvanishing of $L(s,\chi)$ on the line ${\rm Re}(s) = 1$ for all Dirichlet characters $\chi \bmod m$, With this information, let's see how to derive the first Moebius analogue above. When $(a,m) = 1$, $$ \sum_{\substack{n \leq x \\ n \equiv a \bmod m}} \mu(n) = \sum_{n \leq x} \frac{1}{\varphi(m)}\left(\sum_{\chi \bmod m} \overline{\chi}(a)\chi(n)\right)\mu(n), $$ which is $$ \frac{1}{\varphi(m)}\sum_{\chi \bmod m} \overline{\chi}(a)\left(\sum_{n \leq x} \chi(n)\mu(n)\right). $$ To show this is $o(x)$, we show each inner sum is $o(x)$.

The inner sum at $\chi$ is the partial sum of the coefficients of the Dirichlet series $\sum \chi(n)\mu(n)/n^s$, which is $1/L(s,\chi)$ for ${\rm Re}(s) > 1$. The coefficients of this Dirichlet series are bounded in absolute value (by $1$) and $1/L(s,\chi)$ has an analytic continuation to the line ${\rm Re}(s) = 1$ (this is where we use the nonvanishing of all $L(s,\chi)$ on that line, including the pole at $s = 1$ when $\chi$ is the trivial character, so the reciprocal there is analytic at $s=1$ with a zero), so by Newman's Tauberian theorem $$ \frac{1}{x}\sum_{n \leq x} \chi(n)\mu(n) \to {\rm Res}_{s=1} \frac{1}{L(s,\chi)} = 0. $$ The convergence of $\sum_{n \equiv a \bmod m} \mu(n)/n$ when $(a,m) = 1$ is proved by a similar approach: reduce the task to showing for all Dirichlet characters $\chi \bmod m$ that $\sum_{n \leq x, n \equiv a \bmod m} \chi(n)\mu(n)/n$ converges as $x \to \infty$.

$\endgroup$
1
  • $\begingroup$ The main result of the paper "On Some infinite series involving Arithmetic functions II" by Davenport also seems interesting, where he considers additive twists of partial sums of the Möbius function. I guess one can then take linear combinations and arrive at sharper bounds than what I had mentioned in the question, that is bounds which have an arbitrary power of $\log$ in the denominator. $\endgroup$ Jul 18 at 5:12

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.