{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Statement\n\nLet $(a_n)_{n \\ge 0}$ be real numbers with $a_n \\ge 0$ for all $n$.\nAssume that the [[power series|0VP56B4EMX]] $\\sum_{n \\ge 0} a_n x^n$ converges for $0 \\le x \u003c 1$ and that\n$$\\sum_{n \\ge 0} a_n x^n \\sim \\frac{1}{1 - x} \\quad (x \\to 1^-),$$\nwhere $\\sim$ denotes [[asymptotic equivalence|0QR92XBV6H]].\nThen\n$$\\sum_{k=0}^{n} a_k \\sim n \\quad (n \\to \\infty).$$\n\n## Remarks\n\nKaramata's proof shows first that $(1 - x) \\sum_{n} a_n x^n P(x^n) \\to \\int_0^1 P(t) \\, dt$ for every polynomial $P$, by checking it on monomials $P(t) = t^m$.\nThen it approximates the function $g$ with $g(t) = 1/t$ on $[e^{-1}, 1]$ and $g(t) = 0$ on $[0, e^{-1})$ from above and below by polynomials with close integrals.\nPositivity of $a_n$ makes these approximations compatible with the sums.\nTaking $x = e^{-1/n}$, one has $x^k g(x^k) = 1$ for $k \\le n$ and $0$ for $k \u003e n$, which isolates the partial sum $\\sum_{k \\le n} a_k$.\n\nExample: for $a_n = 1 + (-1)^n$, one has $\\sum a_n x^n = \\frac{1}{1-x} + \\frac{1}{1+x} \\sim \\frac{1}{1-x}$, and indeed $\\sum_{k \\le n} a_k$ is $n + 2$ or $n + 1$.\n\nPositivity is needed.\nFor $a_n = 1 + (-1)^n n$, one has $\\sum a_n x^n = \\frac{1}{1-x} - \\frac{x}{(1+x)^2} \\sim \\frac{1}{1-x}$.\nBut $\\sum_{k=0}^{n} (-1)^k k$ is $\\frac{n}{2}$ for even $n$ and $-\\frac{n+1}{2}$ for odd $n$, so the partial sums $\\sum_{k \\le n} a_k$ are close to $\\frac{3n}{2}$ or to $\\frac{n}{2}$ and are not $\\sim n$.\n\n## Explanations\n\n### L1 human\n\nKnowing how a power series with nonnegative coefficients blows up as $x \\to 1$ tells you how fast its coefficients add up.\nIf the power series behaves like $\\frac{1}{1-x}$, then the coefficients behave on average like $1$, so their first $n$ add up to about $n$.\nThe statement is false without the sign condition, since cancellations can hide in the power series.\n\n### L2 human\n\nThis theorem is the prototype of Tauberian theorems with positivity conditions, where no condition on the size of single terms is needed.\nIn terms of $x = e^{-s}$ it says that $\\sum a_n e^{-sn} \\sim \\frac1s$ as $s \\to 0^+$ implies $\\sum_{k \\le n} a_k \\sim n$; [[Karamata's Tauberian theorem|DBEGF38V9H]] extends it to Laplace–Stieltjes transforms with regularly varying growth.\nThe same method of proof gives [[Littlewood's Tauberian theorem|Z35TD4ZZRD]], and the theorem applies to counting problems in number theory where generating functions are easier to estimate than partial sums.\n",
  "commit": "5fc425f97f1f6b11da5892c2948ef8c5d8805665",
  "front": {
    "about": [
      "D6XWK1NRY1"
    ],
    "kind": "statement",
    "msc": [
      "40E05",
      "40G10"
    ],
    "name": "Hardy–Littlewood Tauberian theorem",
    "role": "theorem",
    "same_as": {
      "wikidata": "Q5656673",
      "wikipedia": "Hardy–Littlewood Tauberian theorem"
    },
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Hardy–Littlewood_Tauberian_theorem"
      }
    ],
    "tag": "N42JDBT0VA",
    "uses": [
      "D6XWK1NRY1",
      "0VP56B4EMX",
      "0QR92XBV6H"
    ]
  },
  "kind": "statement",
  "link": "https://math2.tamnd.com/node/N42JDBT0VA/",
  "links": [
    {
      "field": "about",
      "target": "D6XWK1NRY1"
    },
    {
      "field": "uses",
      "target": "D6XWK1NRY1"
    },
    {
      "field": "uses",
      "target": "0VP56B4EMX"
    },
    {
      "field": "uses",
      "target": "0QR92XBV6H"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/N42JDBT0VA.md",
  "msc": [
    "40E05",
    "40G10"
  ],
  "name": "Hardy–Littlewood Tauberian theorem",
  "prov": null,
  "role": "theorem",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/N42/N42JDBT0VA.md",
  "status": null,
  "tag": "N42JDBT0VA",
  "uri": "m2://N42JDBT0VA",
  "url": "https://math2.tamnd.com/wiki/hardy-littlewood-tauberian-theorem/",
  "writing": {
    "tag": "N42JDBT0VA",
    "kind": "statement",
    "level": "W0",
    "next": [
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      }
    ]
  }
}