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CorollaryStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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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.

For a series of real numbers, unconditional convergence and absolute convergence are the same property

Statement

Let (ak)(a_k) be a sequence of reals. The following are equivalent.

  1. ak\sum a_k converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index).
  2. ak\sum a_k converges unconditionally (Rearrangement of a series along a bijection of N\mathbb{N}, and unconditional convergence).
  3. ak\sum a_k converges and every rearrangement of it converges, with no requirement that the sums agree.

So over R\mathbb{R} there is nothing between absolute and conditional convergence: a convergent series either may be reordered freely, sum and all, or else has a rearrangement that fails to converge at all.

This is a statement about R\mathbb{R}, and nothing here says how much of it survives elsewhere. Whether the equivalence of 1 and 2 holds for series of vectors is a question this library cannot pose at this point in the reading order, since it has no notion of a convergent series of vectors; it is raised, and left open, in The same question in Rd\mathbb{R}^d: what the set of rearrangement sums looks like, and why that answer is not reachable at this point in the reading order. No claim about any space other than R\mathbb{R} is made or used here.

Facts & Assumptions

Given: A sequence (ak)(a_k) of reals.

[L1]

An absolutely convergent series converges unconditionally: every rearrangement converges, to the same sum (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum).

[L2]

If ak\sum a_k converges conditionally then for every αβ\alpha \le \beta in the extended reals there is a rearrangement whose partial sums have those as limit inferior and limit superior; in particular there is one whose partial sums diverge to ++\infty (The Riemann series theorem: a conditionally convergent real series has, for every cRc \in \mathbb{R}, a rearrangement with sum cc, and rearrangements diverging to ++\infty, to -\infty, and oscillating with any prescribed lim inflim sup\liminf \le \limsup in R\overline{\mathbb{R}}).

[L3]

Unconditional convergence means: the series converges, and every rearrangement converges to the same sum (Rearrangement of a series along a bijection of N\mathbb{N}, and unconditional convergence).

[L4]

A series converges absolutely when ak\sum |a_k| converges, and conditionally when it converges while ak\sum |a_k| does not; a convergent series is exactly one of the two (Absolutely convergent and conditionally convergent series, and the general starting index).

[L5]

A sequence diverging to ++\infty does not converge: if xn+x_n \to +\infty and also xnLx_n \to L, then eventually xn>L+1x_n > L + 1 and eventually xnL<1|x_n - L| < 1, which are incompatible (Divergence to ++\infty and to -\infty, Limits and Cauchy sequences of reals, Series, partial sums, convergence and the sum, divergence, and the tail series).

Proof

technique · direct
1.1

Assume 1. Then by [L1] the series converges and every rearrangement converges to the same sum, which is 2.

L1L3
1.2

Assume 2. Then in particular the series converges and every rearrangement converges, which is 3.

L3
1.3

Assume 3, and suppose ak\sum a_k did not converge absolutely. Since it converges, it would then converge conditionally.

L4
2.1

In that situation [L2] supplies a bijection σ\sigma of N\mathbb{N} for which the partial sums of aσ(k)\sum a_{\sigma(k)} diverge to ++\infty, and such a series does not converge; this contradicts the assumption that every rearrangement converges.

step 1.3L2L5
3.1

Hence under 3 the series converges absolutely, which is 1.

step 1.3step 2.1L4
4.1

The implications 1 to 2, 2 to 3 and 3 to 1 close the cycle, so the three statements are equivalent.

step 1.1step 1.2step 3.1

Remarks

Depends on

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Sources