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

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Mertens' first theorem for primes

Statement

For every real $x \geq 2$ ,

\sum_{p \leq x} \frac{\log p}{p} = \log x + O (1) .

Facts & Assumptions

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

[L1]

The weighted von Mangoldt harmonic sum satisfies $\sum_{n \leq x} \frac{Λ (n)}{n} = \log x + O (1)$ (The von Mangoldt harmonic sum is log x plus O(1)).

[L2]

The von Mangoldt function is $Λ (p^{k}) = \log p$ on prime powers and $0$ otherwise (The von Mangoldt function, Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L3]

The real $p$ -series $\sum_{m \geq 1} m^{- 3 / 2}$ converges, and comparison for nonnegative series is valid (The p-series for a real exponent p converges exactly when p is greater than one, If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ).

[L4]

The logarithm is increasing and satisfies $\log (u v) = \log u + \log v$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

Proof

technique · direct

1.1

By [L2], $\sum_{n \leq x} \frac{Λ (n)}{n} = \sum_{p \leq x} \frac{\log p}{p} + \sum_{\begin{array}{c} p^{k} \leq x \\ k \geq 2 \end{array}} \frac{\log p}{p^{k}} .$ So it is enough to show that the prime-power tail is bounded independently of $x$ .

L2givenalgebra

2.1

For $t \geq 16$ , define $h (t) : = \sqrt{t} - \log t$ . By [L4], $h^{'} (t) = \frac{1}{2 \sqrt{t}} - \frac{1}{t} = \frac{\sqrt{t} - 2}{2 t} \geq 0,$ so $h$ is increasing on $[16, \infty)$ . Since $h (16) = 4 - \log 16 > 0$ , we obtain $\log p \leq \sqrt{p}$ for every prime $p \geq 16$ . Hence $\sum_{k \geq 2} \frac{\log p}{p^{k}} \leq \frac{\log p}{p^{2}} \sum_{j \geq 0} \frac{1}{p^{j}} \leq \frac{2 \log p}{p^{2}} \leq \frac{2}{p^{3 / 2}}$ for every prime $p \geq 16$ . The finitely many primes $p < 16$ contribute only a constant, so [L3] shows that $\sum_{p^{k} \leq x, k \geq 2} \frac{\log p}{p^{k}} = O (1) .$

L3L4step 1.1algebra

3.1

Combine step 2.1 with [L1]: $\log x + O (1) = \sum_{n \leq x} \frac{Λ (n)}{n} = \sum_{p \leq x} \frac{\log p}{p} + O (1) .$ Therefore $\sum_{p \leq x} \frac{\log p}{p} = \log x + O (1) .$

L1step 1.1step 2.1algebra∎

Depends on

Used by

Numerics for the first and second Mertens theorems Example
Mertens' second theorem for primes Theorem
Mertens' third theorem for primes Theorem

Dependency tree · two levels

51 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)
Victor Shoup, A Computational Introduction to Number Theory and Algebra, Version 2 (standard reference, not scraped)