CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The sum of the reciprocals of the primes diverges

Statement

The series

$\sum_{p} \frac{1}{p}$

diverges.

Facts & Assumptions

Given: The reciprocal-prime partial sums.

[L1]

Mertens' second theorem gives $\sum_{p \leq x} \frac{1}{p} = \log \log x + B_{1} + O (1 / \log x)$ (Mertens' second theorem for primes).

Proof

technique · direct

1.1L1

By [L1], the partial sums differ from $\log \log x$ by a bounded quantity as $x \to \infty$ .

2.1step 1.1algebra∎

Since $\log \log x \to \infty$ , the partial sums $\sum_{p \leq x} \frac{1}{p}$ also tend to infinity. Therefore the prime reciprocal series diverges.

Depends on

Mertens' second theorem for primes

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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)