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

arXiv eprints
 Pub Date:
 August 2009
 DOI:
 10.48550/arXiv.0908.4323
 arXiv:
 arXiv:0908.4323
 Bibcode:
 2009arXiv0908.4323T
 Keywords:

 Mathematics  Number Theory;
 11A25
 EPrint:
 7 pages, no figures. To appear, Bull. Aust. Math. Soc. Minor corrections