Alphabeta Math
DefinitionDefinition: AI-adaptedProof: Not applicableSession-authored (Fable 5 assisted)judge pass (z-ai/glm-5.2)audited 2026-07-27
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Absolutely convergent and conditionally convergent series, and the general starting index

Definition

Let (ak)(a_k) be a sequence of reals, with series ak\sum a_k and partial sums sn=k<naks_n = \sum_{k<n} a_k as in Series, partial sums, convergence and the sum, divergence, and the tail series, and let x|x| be the absolute value (Absolute value in an ordered field).

Absolute convergence. The series ak\sum a_k converges absolutely when the series ak\sum |a_k| converges (Series, partial sums, convergence and the sum, divergence, and the tail series). Since ak0|a_k| \ge 0 for every kk (Basic properties of the absolute value), this is a statement about a series of nonnegative terms.

Conditional convergence. The series ak\sum a_k converges conditionally when it converges (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals) and does not converge absolutely.

So a convergent series is exactly one of the two: absolutely convergent or conditionally convergent, according as ak\sum |a_k| converges or not.

One implication is already proved, and is not reproved anywhere on this page. If ak\sum |a_k| converges then ak\sum a_k converges states that if ak\sum |a_k| converges then ak\sum a_k converges. That lemma was coined and proved on the previous page of this track, where the root and ratio tests need it; this page names it and builds on it. In particular an absolutely convergent series is a convergent series, so the two words above really do partition the convergent series, and "conditionally convergent" is not vacuous by accident: the alternating harmonic series is a witness, and the witness is exhibited in FALSE: every convergent series converges absolutely.

General starting index. Let mNm \in \mathbb{N} and let (ak)km(a_k)_{k \ge m} be a family from mm (Series, partial sums, convergence and the sum, divergence, and the tail series). The series kmak\sum_{k \ge m} a_k converges absolutely when kmak\sum_{k \ge m} |a_k| converges, and converges conditionally when it converges and does not converge absolutely. By Series, partial sums, convergence and the sum, divergence, and the tail series both statements are the corresponding statements for the shifted sequence jaj+mj \mapsto a_{j+m}, so nothing new is being defined and every result below transfers to a general starting index in the same way, exactly as If ak\sum |a_k| converges then ak\sum a_k converges already records for the one implication it proves.

Remarks

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 39 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources