CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The average order of sigma is (pi squared over 6)n

Statement

The arithmetic function $n \mapsto (π^{2} / 6) n$ is an average order of $σ$ .

Facts & Assumptions

Given: A real $x \geq 1$ and $N : = ⌊ x ⌋$ .

Proof

technique · direct

1.1

The comparison sum is exact: $\sum_{n \leq x} \frac{π^{2}}{6} n = \frac{π^{2}}{6} \sum_{n = 1}^{N} n = \frac{π^{2}}{12} N (N + 1) = \frac{π^{2}}{12} x^{2} + O (x) .$

givenalgebra

2.1

By The summatory divisor-sum function is pi squared over 12 times x squared plus O(x log x), $\sum_{n \leq x} σ (n) = π^{2} x^{2} / 12 + O (x \log x)$ . Comparing this with step 1.1 shows that the ratio of the two summatory functions tends to $1$ , and the comparison sum is eventually positive. Therefore Summatory functions and average orders makes $(π^{2} / 6) n$ an average order of $σ$ .

step 1.1givenalgebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

Karl-Dieter Crisman, Number Theory: In Context and Interactive (standard reference, not scraped)