LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Nonprincipal Dirichlet character partial sums are bounded

Statement

Let $χ$ be a nonprincipal Dirichlet character modulo $q$ . Then for every real $x \geq 1$ ,

$∣ \sum_{1 \leq n \leq x} χ (n) ∣ \leq q .$

Facts & Assumptions

Given: A nonprincipal Dirichlet character $χ$ modulo $q$ and a real $x \geq 1$ .

[L1]

Dirichlet characters are periodic modulo $q$ (Extension by zero is well defined and periodic).

[L2]

The sum of $χ$ over any complete residue system modulo $q$ is $0$ (A nonprincipal character has zero complete sum).

Proof

technique · direct

1.1

Write $⌊ x ⌋ = m q + r$ with integers $m \geq 0$ and $0 \leq r < q$ . By [L1], the sum over $1 \leq n \leq x$ is the sum over $1 \leq n \leq ⌊ x ⌋$ , which breaks into $m$ complete blocks of length $q$ and one terminal block of length $r$ .

L1givenalgebra

2.1

Every complete block contributes $0$ by [L2]. Hence $\sum_{1 \leq n \leq x} χ (n) = \sum_{m q < n \leq m q + r} χ (n)$ . The terminal block has at most $q - 1$ terms, and every term has modulus at most $1$ because character values are either $0$ or roots of unity. Therefore the absolute value of the sum is at most $q - 1 < q$ , so certainly at most $q$ .

L2step 1.1algebra∎

Depends on

Used by

Dependency tree · two levels

5 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

Leonard Tomczak, Analytic Number Theory, Chapter 4 (standard reference, not scraped)
Kiran S. Kedlaya, Notes on Analytic Number Theory, section 3.1 (standard reference, not scraped)