{
  "id": "0908.4323",
  "version": "v5",
  "published": "2009-08-29T06:28:53.000Z",
  "updated": "2009-10-05T20:22:08.000Z",
  "title": "A remark on partial sums involving the Mobius function",
  "authors": [
    "Terence Tao"
  ],
  "comment": "7 pages, no figures. To appear, Bull. Aust. Math. Soc. Minor corrections",
  "categories": [
    "math.NT"
  ],
  "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 $m,x,N,P \\geq 1$.",
  "revisions": [
    {
      "version": "v5",
      "updated": "2009-10-05T20:22:08.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "11A25"
    ],
    "keywords": [
      "partial sums",
      "mobius function",
      "prime number theorem",
      "well-known observation",
      "elementary proof"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 7,
      "language": "en",
      "license": "arXiv",
      "status": "editable",
      "adsabs": "2009arXiv0908.4323T"
    }
  }
}