FormalConjectures/ErdosProblems/314.lean

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L4DEL
Lean4 Domain Explanation Language

Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.

Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

3 declarations, 1 contain sorry, 0 declared as axioms

Erdős Problem 314

References:

Imports (1)
  1. def Erdos314.mMin mMin n is the minimal m such that $\sum_{n\leq k\leq m}\frac{1}{k}\geq 1$; such an m exists for n ≥ 1 since the harmonic series diverges line 39 source

    Source documentation

    mMin n is the minimal m such that $\sum_{n\leq k\leq m}\frac{1}{k}\geq 1$; such an m exists for n ≥ 1 since the harmonic series diverges.

    standard axioms: Classical.choice, Quot.sound, propext what this means

    Statement uses: Nat

    Proof uses: InfSet.sInf, Set.ofPred, LE.le, Rat, OfNat.ofNat, harmonicBlock

    3 instances, coercions and auxiliary constants left out.

    noncomputable def mMin (n : ℕ) : ℕ := sInf {m | 1 ≤ harmonicBlock n m}

    Mentions: harmonicBlock

  2. def Erdos314.epsilon $\epsilon(n) = \sum_{n\leq k\leq m}\frac{1}{k}-1$, the overshoot of the minimal block sum reaching $1$ line 43 source

    Source documentation

    $\epsilon(n) = \sum_{n\leq k\leq m}\frac{1}{k}-1$, the overshoot of the minimal block sum reaching $1$.

    standard axioms: Classical.choice, Quot.sound, propext what this means

    Statement uses: Nat, Real

    Proof uses: HSub.hSub, harmonicBlock, Erdos314.mMin, OfNat.ofNat

    6 instances, coercions and auxiliary constants left out.

    noncomputable def epsilon (n : ℕ) : ℝ := (harmonicBlock n (mMin n) : ℝ) - 1
…

    Mentions: harmonicBlock, Erdos314.mMin

  3. theorem Erdos314.erdos_314 Let $n\geq 1$ and let $m$ be minimal such that $\sum_{n\leq k\leq m}\frac{1}{k}\geq 1$ sorry line 58 source

    Source documentation

    Let $n\geq 1$ and let $m$ be minimal such that $\sum_{n\leq k\leq m}\frac{1}{k}\geq 1$. We define $$\epsilon(n) = \sum_{n\leq k\leq m}\frac{1}{k}-1.$$ How small can $\epsilon(n)$ be? Is it true that $$\liminf n^2\epsilon(n)=0?$$

    This is true, and shown by Lim and Steinerberger [LiSt24], who further proved that, for any $\delta>0$, there exist infinitely many $n$ and $m$ such that $$n^2\left\lvert \sum_{n\leq k\leq m}\frac{1}{k}-1\right\rvert\ll \frac{1}{(\log n)^{5/4-\delta}}.$$ Erdős and Graham (and also Lim and Steinerberger) believe that the exponent of $2$ is best possible here, in that $\liminf \epsilon(n) n^{2+\delta}=\infty$ for all $\delta>0$.

    depends on sorry what this means

    Statement uses: Iff, True, Eq, Real, Filter.liminf, Nat, HMul.hMul, HPow.hPow, OfNat.ofNat, Erdos314.epsilon, Filter.atTop

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    28 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem erdos_314 : answer(True) ↔
    atTop.liminf (fun n : ℕ => (n : ℝ) ^ 2 * epsilon n) = 0 := by
…

    Mentions: Erdos314.epsilon