False statementConstruction: AI-adaptedVerification: AI-generatedverified 2026-09-26 (gpt-6-sol)

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FALSE: if a convergent series in $R^{n}$ does not converge absolutely, then every point of $R^{n}$ is the sum of some rearrangement of it

Statement

False claim: let $n \geq 1$ and let $(x_{k})$ be a sequence in $R^{n}$ whose series converges but does not converge absolutely (Series of vectors in $R^{n}$ , absolute convergence, rearrangement, and the set of rearrangement sums). Then every point of $R^{n}$ is the sum of some rearrangement of $\sum x_{k}$ ; that is, $S (x) = R^{n}$ .

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 $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ : a conditionally convergent real series can be rearranged to any prescribed sum. The claim above is the naive transfer of that theorem to $R^{n}$ by analogy, and the analogy fails at $n = 2$ already.

The witness uses the alternating sequence $ε_{k} = (- 1)^{k}$ of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ and the series $x_{k} = (ε_{k} / ι (k + 1), 0)$ in $R^{2}$ , 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 $(x_{k})$ in $R^{2}$ with $x_{k} = (c_{k}, 0)$ and $c_{k} = ε_{k} / ι (k + 1)$ , where $ε_{k} = (- 1)^{k}$ is the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ .

[A1]

The refuted claim, instantiated at $n = 2$ and this $(x_{k})$ : every point of $R^{2}$ , in particular $(0, 1)$ , is the sum of some rearrangement of $\sum x_{k}$ .

[L1]

The real series $\sum_{k} c_{k}$ converges by The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ applied to the nonincreasing null sequence $1 / ι (k + 1)$ . Thus $\sum x_{k}$ converges componentwise, with sum $s = (S, 0)$ ; it does not converge absolutely, since $∥ x_{k} ∥_{2} = 1 / ι (k + 1)$ and the harmonic series diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , For $n \geq 1$ a sequence in $R^{n}$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $R^{n}$ is complete in every norm, Absolutely convergent and conditionally convergent series, and the general starting index, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L2]

A rearrangement of $\sum x_{k}$ is $\sum x_{σ (k)}$ for a bijection $σ$ of $N$ , and $S (x)$ is the set of its sums (Series of vectors in $R^{n}$ , absolute convergence, rearrangement, and the set of rearrangement sums, Injection, surjection, bijection).

[L3]

Convergence in $R^{2}$ is componentwise, partial sums are computed coordinatewise, and a limit in a metric space is unique (For $n \geq 1$ a sequence in $R^{n}$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $R^{n}$ is complete in every norm clause 1, The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ clause 1, A sequence in a metric space has at most one limit, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L4]

The containment theorem gives $S (x) \subseteq s + Γ^{⊥}$ (The set of rearrangement sums of a convergent series in $R^{n}$ is a nonempty subset of the affine subspace $s + Γ^{⊥}$ , The subspace $Γ$ of directions along which a series converges absolutely, and its orthogonal complement $Γ^{⊥}$ ).

Refutation

technique · direct

1.1L2L3

The second coordinate of every term $x_{k}$ is $0$ , so for every bijection $σ$ of $N$ the second coordinate of every partial sum $\sum_{j < N} x_{σ (j)}$ is the finite sum of zeros, namely $0$ .

1.2L1

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

2.1step 1.1L3

If a rearrangement $\sum x_{σ (k)}$ converges to a point $t \in R^{2}$ , then by componentwise convergence its second coordinate sequence, constantly $0$ by step 1.1, converges to $t_{1}$ ; a constant sequence converges to its value and limits are unique, so $t_{1} = 0$ .

3.1step 2.1L2

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

4.1step 1.2step 3.1A1

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

5.1L1L4given∎

For $a = (a_{0}, a_{1})$ , $∣ ⟨ a, x_{k} ⟩ ∣ = ∣ a_{0} ∣ / ι (k + 1)$ . By harmonic divergence, this series converges exactly when $a_{0} = 0$ : if $a_{0} \neq 0$ , multiplication by $1 / ∣ a_{0} ∣$ would otherwise make the harmonic series converge. Thus $Γ$ is the vertical axis, and its orthogonal complement $Γ^{⊥}$ is the horizontal axis. The containment theorem therefore also puts every rearrangement sum in the proper affine line $s + Γ^{⊥}$ .

Remarks

The refutation uses only the containment half. Step 2.1 is an elementary argument about the second coordinate and needs nothing beyond componentwise convergence; step 5.1 explains it through The set of rearrangement sums of a convergent series in $R^{n}$ is a nonempty subset of the affine subspace $s + Γ^{⊥}$ , which proves $S (x) \subseteq s + Γ^{⊥}$ and nothing more. No statement about the reverse inclusion is used here, and none is asserted.
Why $n = 1$ is genuinely different. For a conditionally convergent real series $Γ = {0}$ , so $Γ^{⊥}$ is the whole line and the containment says nothing; the space simply has no proper subspace for the rearrangement sums to be trapped in other than ${0}$ . From $n = 2$ on there is room, and this witness uses it.
What a correct general statement would have to look like. The affine subspace $s + Γ^{⊥}$ is an upper bound for $S (x)$ , and the two extremes are both realised: it is a single point when the series converges absolutely (An absolutely convergent series in $R^{n}$ converges, and every rearrangement converges to the same sum), and it is the whole line in the one-dimensional conditionally convergent case (The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ ). What happens between those extremes for $n \geq 2$ is not settled in this library, and the present item settles only that "everything" is the wrong answer.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

122 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Riemann series theorem (Wikipedia) (standard reference, not scraped)
Levy-Steinitz theorem (Wikipedia) (standard reference, not scraped)
T. Banakh, A Simple Inductive Proof of the Levy-Steinitz Theorem (standard reference, not scraped)

FALSE: if a convergent series in RnRn does not converge absolutely, then every point of RnRn is the sum of some rearrangement of it