Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge 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.

FALSE: every rearrangement of a convergent series converges, and to the same sum

Statement

False claim: for every sequence (ak)(a_k) of reals whose series converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and every bijection σ:NN\sigma : \mathbb{N} \to \mathbb{N}, the rearranged series aσ(k)\sum a_{\sigma(k)} (Rearrangement of a series along a bijection of N\mathbb{N}, and unconditional convergence) converges, with the same sum.

What is true is that hypothesis: the claim holds for absolutely convergent series, and that is Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum. Dropping "absolutely" makes it false in both of its assertions at once, and the same witness refutes both.

Let (εj)(\varepsilon_j) be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing e,oe, o with N\mathbb{N} their disjoint union, and the unique (sk)(s_k) with s0=1s_0 = 1, sσ(k)=sks_{\sigma(k)} = -s_k, which satisfies sk=1|s_k| = 1, se1s \circ e \equiv 1 and so1s \circ o \equiv -1) and put aj:=εj/ι(j+1)a_j := \varepsilon_j/\iota(j+1), the alternating harmonic series. It converges, by the alternating series test, and does not converge absolutely, its series of absolute values being the harmonic series (For rational p>0p > 0, 1/kp\sum 1/k^p converges iff p>1p > 1). So it converges conditionally, and 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}} applies to it.

Facts & Assumptions

Given: The alternating sequence (εj)(\varepsilon_j), the sequence bj:=1/ι(j+1)b_j := 1/\iota(j+1), and aj:=εjbja_j := \varepsilon_j b_j, whose series is the alternating harmonic series.

[A1]

The refuted claim: for every convergent series of reals and every bijection of N\mathbb{N}, the rearranged series converges with the same sum.

[L2]

The canonical naturals ι(n)\iota(n) are positive for n1n \ge 1 and strictly increasing; if 0<u<v0 < u < v then 0<1/v<1/u0 < 1/v < 1/u; and for every real ε>0\varepsilon > 0 there is n1n \ge 1 with 1/ι(n)<ε1/\iota(n) < \varepsilon (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon).

[L4]

k11/kp\sum_{k\ge1} 1/k^{p} converges if and only if p>1p > 1, with ι(k)1=ι(k)\iota(k)^{1} = \iota(k); and k1xk\sum_{k \ge 1} x_k is the series of jxj+1j \mapsto x_{j+1} (For rational p>0p > 0, 1/kp\sum 1/k^p converges iff p>1p > 1, Rational powers ara^r of a positive base, Existence and uniqueness of nn-th roots: a unique a1/n0a^{1/n} \ge 0 with (a1/n)n=a(a^{1/n})^n = a, Integer powers ama^m, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L5]

Absolute value: xy=xy|xy| = |x|\,|y| (Basic properties of the absolute value).

Refutation

technique · direct
1.1

The sequence (bj)(b_j) is positive, nonincreasing and converges to 00: positivity and monotonicity from 0<ι(j+1)<ι(j+2)0 < \iota(j+1) < \iota(j+2), and convergence because, given a rational ε>0\varepsilon > 0, an n1n \ge 1 with 1/ι(n)<ε1/\iota(n) < \varepsilon satisfies bj1/ι(n)<εb_j \le 1/\iota(n) < \varepsilon for every jnj \ge n.

givenL2
2.1

By the alternating series test aj\sum a_j converges; write SS for its sum.

step 1.1L3
2.2

For every jj, aj=εjbj=1/ι(j+1)|a_j| = |\varepsilon_j| b_j = 1/\iota(j+1), and j1/ι(j+1)\sum_j 1/\iota(j+1) is the pp-series k11/k\sum_{k\ge1}1/k at p=1p = 1, which diverges.

step 1.1L1L4L5
3.1

So aj\sum a_j converges conditionally.

step 2.1step 2.2L6
4.1

By the Riemann series theorem there is a bijection σ\sigma of N\mathbb{N} with aσ(k)\sum a_{\sigma(k)} convergent of sum S+1S + 1, a number different from SS.

step 3.1L7
4.2

By the same theorem there is a bijection τ\tau of N\mathbb{N} for which the partial sums of aτ(k)\sum a_{\tau(k)} diverge to ++\infty, so that rearranged series does not converge at all.

step 3.1L7
5.1

The claim [A1] therefore fails twice over for the alternating harmonic series: once in its assertion that the sum is preserved, by step 4.1, and once in its assertion that the rearranged series converges, by step 4.2.

step 4.1step 4.2A1
6.1

The claim is false. What is true is the same statement with "converges" strengthened to "converges absolutely" in the hypothesis.

step 5.1A1L8

Remarks

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 139 results over 32 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