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False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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FALSE: if a convergent series in Rn does not converge absolutely, then every point of Rn is the sum of some rearrangement of it

Statement

False claim: let n≥1 and let (xk) be a sequence in Rn whose series converges but does not converge absolutely (Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums). Then every point of Rn is the sum of some rearrangement of ∑xk; that is, S(x)=Rn.

Where the claim comes from. For n=1 it is true, and it is the published 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‾: a conditionally convergent real series can be rearranged to any prescribed sum. The claim above is the naive transfer of that theorem to Rn by analogy, and the analogy fails at n=2 already.

The witness is the series of A convergent series in R2 with Γ a line and Γ⊥ a line, computed from the definition: xk=(εk/ι(k+1), 0) in R2, which converges, does not converge absolutely, and has no rearrangement sum off the horizontal axis. In particular (0,1) is not a rearrangement sum.

Facts & Assumptions

Given: The sequence (xk) in R2 of A convergent series in R2 with Γ a line and Γ⊥ a line, computed from the definition, with xk=(ck,0) and ck=εk/ι(k+1).

[A1]

The refuted claim, instantiated at n=2 and this (xk): every point of R2, in particular (0,1), is the sum of some rearrangement of ∑xk.

[L2]

A rearrangement of ∑xk is ∑xσ(k) for a bijection σ of N, and S(x) is the set of its sums (Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums, Injection, surjection, bijection).

Refutation

technique · direct
1.1

The second coordinate of every term xk is 0, so for every bijection σ of N the second coordinate of every partial sum ∑j<Nxσ(j) is the finite sum of zeros, namely 0.

L2L3
1.2

The hypotheses of the refuted claim are met by this series: it converges and does not converge absolutely.

L1
2.1

If a rearrangement ∑xσ(k) converges to a point t∈R2, then by componentwise convergence its second coordinate sequence, constantly 0 by step 1.1, converges to t1; a constant sequence converges to its value and limits are unique, so t1=0.

step 1.1L3
3.1

Hence every element of S(x) has second coordinate 0, and (0,1), whose second coordinate is 1≠0, is not a rearrangement sum.

step 2.1L2
4.1

So [A1] fails for a series satisfying the hypotheses of the refuted claim, and the claim is false.

step 1.2step 3.1A1
5.1

The failure is structural rather than accidental: by the containment theorem every rearrangement sum lies in s+Γ⊥, and here Γ⊥ is a line in R2, hence a proper subset, so S(x) cannot be all of R2 whatever else is true of it.

step 3.1L4∎

Remarks

Depends on

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Sources