Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Almost-sure conditional convergence

Example

Assume countable choice and dependent choice. On a probability space carrying independent fair signs ϵn{1,1}, the random harmonic series n1ϵn/n converges almost surely, but n1ϵn/n diverges at every sample point. By comparison the deterministic harmonic series diverges and its alternating version converges.

Facts & Assumptions

[F1]

Kolmogorov three-series theorem: Let (Xn)n1 be independent real random variables and fix A>0. Put Yn=Xn1{XnA}. Then nXn converges almost surely if and only if all three conditions hold: nP(Xn>A)<,nEYn converges in R,nVar(Yn)<. The conditions hold for some A>0 if and only if they hold for every A>0. No moment assumption is imposed on the untruncated variables.

[F2]

Countably many independent copies of a prescribed law exist: Assume countable choice and dependent choice. Every probability measure ν on (S,Σ) is the common law of a countable independent family of S-valued random elements.

[F3]

The p-series for a real exponent p converges exactly when p is greater than one: For every real p, k11kp convergesp>1.

[F4]

The alternating series test: if (bk) is nonincreasing with bk0 then k(1)kbk converges, the sum lies between any two consecutive partial sums, and the error after n terms is at most bn: Let (εk) be the alternating sequence of lem-alternating-sequence, that is the unique sequence of reals with ε0=1 and εk+1=εk, which is what is usually written εk=(1)k; let e and o be its even and odd index maps, so that εej=1, εoj=1, and every natural number is ej for exactly one j or oj for exactly one j. Let (bk) be a sequence of reals that is nonincreasing (def-monotone-sequence) and converges to 0 (def-real-limit); then bk0 for every k. Write tn:=k<nεkbk for the partial sums (def-series). Then: 1. the series εkbk converges; write L for its sum; 2. tejLtoj for every jN, and for every nN the sum L lies between the two consecutive partial sums tn and tn+1; 3. Ltnbn for every nN. Claim 3 is the error bound: the partial sum tn, which uses the n terms ε0b0,,εn1bn1, differs from the sum by at most the first term omitted. Only claim 1 is a corollary of thm-dirichlet-test. Claims 2 and 3 are not: they come from the interlacing of the even-index and odd-index partial sums, and that argument is carried out below rather than smuggled into the Dirichlet estimate, which produces no bracketing at all.

Verification

Given: The construction and assumptions above.

1.1

Under countable choice and dependent choice, use the countable-copy theorem for the fair law on {1,1}. At cutoff A=1, all summands Xn=ϵn/n are retained, including the first one. Their means are zero and their variances are 1/n2, whose sum is finite. Three-series therefore gives almost-sure convergence.

F2F3F1givenalgebra
2.1

At every point ϵn/n=1/n, and the harmonic p-series diverges. Thus on the probability-one convergence event the convergence is conditional. The same p-series test gives deterministic harmonic divergence. Apply the zero-based alternating-series test with bk=1/(k+1) to obtain convergence of n1(1)n1/n; bk decreases to zero and is nonnegative.

F3F4step 1.1algebra

Depends on

Used by

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Sources