LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01

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

Statement

For every real $x \geq 2$ ,

$\sum_{n \leq x} \frac{Λ (n)}{n} = \log x + O (1) .$

Facts & Assumptions

Given: A real number $x \geq 2$ .

[L1]

The von Mangoldt divisor identity is $\log m = \sum_{\begin{array}{c} d ∣ m \\ d > 0 \end{array}} Λ (d)$ for every integer $m \geq 1$ (The divisor sum of von Mangoldt is the arithmetic-function logarithm, The von Mangoldt function).

[L2]

The summatory logarithm satisfies $\sum_{m \leq x} \log m = x \log x - x + O (\log x)$ (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 $\sum_{d \leq x} Λ (d) = ψ (x) = O (x)$ (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

1.1L1givenconstructalgebra

Summing [L1] over the positive integers $1 \leq m \leq x$ and reversing the finite order of summation gives $\sum_{1 \leq m \leq x} \log m = \sum_{1 \leq d \leq x} Λ (d) ⌊ \frac{x}{d} ⌋ .$ Write $S (x) : = \sum_{1 \leq d \leq x} \frac{Λ (d)}{d} .$ Since $⌊ x / d ⌋ = x / d + O (1)$ , we obtain $\sum_{1 \leq m \leq x} \log m = x S (x) + O ⁣ (\sum_{1 \leq d \leq x} Λ (d)) .$

2.1L3step 1.1algebra

By [L3], the error term in step 1.1 is $O (x)$ . Therefore $\sum_{1 \leq m \leq x} \log m = x S (x) + O (x) .$

3.1L2step 2.1algebra∎

Substitute the asymptotic from [L2] into step 2.1: $x \log x - x + O (\log x) = x S (x) + O (x) .$ 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 $\sum_{n \leq x} \frac{Λ (n)}{n} = \log x + O (1) .$

Depends on

Used by

Mertens' first theorem for primes Theorem

Dependency tree · two levels

24 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Leo Goldmakher, A Quick Proof of Mertens' Theorem (standard reference, not scraped)
MIT 18.785 Number Theory I, Fall 2021, Problem Set 9 (standard reference, not scraped)