Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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.

Abel summation by parts for complex coefficients and their partial sums

Statement

Let a0,,aNC, let sn=k=0nak, and put s1=0. For complex weights b0,,bN, n=0Nanbn=sNbN+n=0N1sn(bnbn+1). More generally, for 0pq, n=pqanbn=sqbqsp1bp+n=pq1sn(bnbn+1).

Facts & Assumptions

Given: Finite complex sequences (an) and (bn) with partial sums sn.

[L1]

Complex partial sums are the finite sums in the additive monoid of C (Complex series, absolute convergence, complex power series, and radius of convergence).

[L2]

Finite products, read additively, have the empty and one-term conventions and obey the recursion defining finite sums (The product g0g1gn1 of a finite list in a monoid, by recursion, with the empty product (n=0) equal to the identity).

Proof

technique · direct
1.1

Since an=snsn1, distributivity gives n=pqanbn=n=pqsnbnn=pqsn1bn.

L1algebra
2.1

Shift the second finite index and collect equal sn terms; the endpoints are sqbq and sp1bp, while the interior terms are sn(bnbn+1).

step 1.1L2algebra
3.1

This is the tail identity. Taking p=0 and s1=0 gives the first display; when p=q the interior sum is empty and the identity remains valid.

step 2.1L2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 69 results over 23 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