Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-07-31
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.

Tauber's theorem: an Abel-summable series with ι(n+1)an→0 converges ordinarily to its Abel sum

Statement

Let ∑n≥0an be Abel summable to s. If

ι(n+1)an⟶0,

then its ordinary partial sums converge to s.

Facts & Assumptions

Given: The Abel sum A(x):=∑n≥0anxn→s as x↑1 and the stated Tauber condition.

[L1]

The block lemma supplies uniform bounds for the weighted middle and tail when xN:=1−1/ι(N+1) (If ι(n+1)an→0, short multiplicative blocks of the coefficients have uniformly small sums).

[L2]

The Archimedean reciprocal property gives a reciprocal below every positive tolerance. Canonical naturals increase and reciprocation reverses positive order, so every later reciprocal remains below that tolerance; hence 1/ι(N+1)→0 and xN↑1. Abel summability then gives A(xN)→s (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, Abel summability by lim⁡x↑1∑anxn and Cesaro summability by the Cesaro means of the partial sums, Limits and Cauchy sequences of reals).

Proof

technique · direct
1.1

Write SN:=∑n=0Nan. For N≥1, one has SN−A(xN)=∑n=0Nan(1−xNn)−∑n>NanxNn.

givenalgebra
2.1

Given ε>0, choose N0 from [L1]. The part of the first sum with n<N0 tends to 0 because it is finite and xN→1; the remaining part and the tail have absolute value at most ε each by [L1].

step 1.1L1choose
3.1

Hence SN−A(xN)→0. Since A(xN)→s by [L2], it follows that SN→s.

step 2.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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