Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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FALSE: every rearrangement of a convergent series converges, and to the same sum

Statement

False claim: for every sequence (ak) of reals whose series converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and every bijection σ:N→N, the rearranged series ∑aσ(k) (Rearrangement of a series along a bijection of 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) be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing e,o with N their disjoint union, and the unique (sk) with s0=1, sσ(k)=−sk, which satisfies ∣sk∣=1, s∘e≡1 and s∘o≡−1) and put aj:=εj/ι(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>0, ∑1/kp converges iff p>1). So it converges conditionally, and The Riemann series theorem: a conditionally convergent real series has, for every c∈R, a rearrangement with sum c, and rearrangements diverging to +∞, to −∞, and oscillating with any prescribed lim inf⁡≤lim sup⁡ in R‾ applies to it.

Facts & Assumptions

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

[A1]

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

[L2]

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

[L5]

Absolute value: ∣xy∣=∣x∣ ∣y∣ (Basic properties of the absolute value).

Refutation

technique · direct
1.1givenL2

The sequence (bj) is positive, nonincreasing and converges to 0: positivity and monotonicity from 0<ι(j+1)<ι(j+2), and convergence because, given a rational ε>0, an n≥1 with 1/ι(n)<ε satisfies bj≤1/ι(n)<ε for every j≥n.

2.1step 1.1L3

By the alternating series test ∑aj converges; write S for its sum.

2.2step 1.1L1L4L5

For every j, ∣aj∣=∣εj∣bj=1/ι(j+1), and ∑j1/ι(j+1) is the p-series ∑k≥11/k at p=1, which diverges.

3.1step 2.1step 2.2L6

So ∑aj converges conditionally.

4.1step 3.1L7

By the Riemann series theorem there is a bijection σ of N with ∑aσ(k) convergent of sum S+1, a number different from S.

4.2step 3.1L7

By the same theorem there is a bijection τ of N for which the partial sums of ∑aτ(k) diverge to +∞, so that rearranged series does not converge at all.

5.1step 4.1step 4.2A1

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.

6.1step 5.1A1L8∎

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

Remarks

Depends on

Dependency tree · 2 levels

Sources