Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The von Mangoldt harmonic sum is log x plus O(1)

Statement

For every real x≥2,

∑n≤xΛ(n)n=log⁡x+O(1).

Facts & Assumptions

Given: A real number x≥2.

[L1]

The von Mangoldt divisor identity is

log⁡m=∑d∣md>0Λ(d)

for every integer m≥1 (The divisor sum of von Mangoldt is the arithmetic-function logarithm, The von Mangoldt function).

[L2]

The summatory logarithm satisfies

∑m≤xlog⁡m=xlog⁡x−x+O(log⁡x)

(The summatory logarithm is x log x minus x plus O(log x)).

[L3]

The prime-power expansion of ψ together with the comparison lemma and Chebyshev's theta bounds imply

∑d≤xΛ(d)=ψ(x)=O(x)

(Chebyshev's psi function, Prime-power expansion of Chebyshev's psi function, Psi and theta differ by at most a square-root term, Chebyshev's theta function has linear lower and upper bounds).

Proof

technique · direct
1.1

Summing [L1] over the positive integers 1≤m≤x and reversing the finite order of summation gives ∑1≤m≤xlog⁡m=∑1≤d≤xΛ(d)⌊xd⌋. Write S(x):=∑1≤d≤xΛ(d)d. Since ⌊x/d⌋=x/d+O(1), we obtain ∑1≤m≤xlog⁡m=xS(x)+O ⁣(∑1≤d≤xΛ(d)).

L1givenconstructalgebra
2.1

By [L3], the error term in step 1.1 is O(x). Therefore ∑1≤m≤xlog⁡m=xS(x)+O(x).

L3step 1.1algebra
3.1

Substitute the asymptotic from [L2] into step 2.1: xlog⁡x−x+O(log⁡x)=xS(x)+O(x). After moving the O(x) term to the left and dividing by x, this becomes S(x)=log⁡x+O(1). That is exactly the claimed estimate ∑n≤xΛ(n)n=log⁡x+O(1).

L2step 2.1algebra∎

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