A remark on partial sums involving the Mobius function

TL;DR

This paper provides an elementary proof that partial sums of the Möbius function over multiplicative subsemigroups are bounded and converge to a product over primes, even with complex zeros of related zeta functions.

Contribution

It introduces a new elementary proof for the boundedness and convergence of Möbius sums over arbitrary prime-generated subsemigroups, extending known results.

Findings

Partial sums are bounded by 1 in magnitude.

Sums converge to the product over primes in P.

Convergence holds despite non-trivial zeros of the associated zeta function.

Abstract

Let $<\P > \subset \N$ be a multiplicative subsemigroup of the natural numbers $\N = \{1,2,3,...\}$ generated by an arbitrary set $\P$ of primes (finite or infinite). We given an elementary proof that the partial sums $\sum_{n \in < \P >: n \leq x} \frac{\mu(n)}{n}$ are bounded in magnitude by 1. With the aid of the prime number theorem, we also show that these sums converge to $\prod_{p \in \P} (1 - \frac{1}{p})$ (the case when $\P$ is all the primes is a well-known observation of Landau). Interestingly, this convergence holds even in the presence of non-trivial zeroes and poles of the associated zeta function $\zeta_\P(s) := \prod_{p \in \P} (1-\frac{1}{p^s})^{-1}$ on the line $\{\Re(s)=1\}$. As equivalent forms of the first inequality, we have $|\sum_{n \leq x: (n,P)=1} \frac{\mu(n)}{n}| \leq 1$, $|\sum_{n|N: n \leq x} \frac{\mu(n)}{n}| \leq 1$, and $|\sum_{n \leq x} \frac{\mu(mn)}{n}| \leq 1$ for all $m,x,N,P \geq 1$.