CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck 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.

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

Statement

Let $(a_{k})$ be a sequence of reals. The following are equivalent.

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

So over $R$ 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 $R$ , 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 $R^{d}$ : 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 $R$ is made or used here.

Facts & Assumptions

Given: A sequence $(a_{k})$ 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 $\sum a_{k}$ converges conditionally then for every $α \leq β$ 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 $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ ).

[L3]

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

[L4]

A series converges absolutely when $\sum ∣ a_{k} ∣$ converges, and conditionally when it converges while $\sum ∣ a_{k} ∣$ 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 $x_{n} \to + \infty$ and also $x_{n} \to L$ , then eventually $x_{n} > L + 1$ and eventually $∣ x_{n} - L ∣ < 1$ , 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

1.1L1L3

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

1.2L3

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

1.3L4

Assume 3, and suppose $\sum a_{k}$ did not converge absolutely. Since it converges, it would then converge conditionally.

2.1step 1.3L2L5

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

3.1step 1.3step 2.1L4

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

4.1step 1.1step 1.2step 3.1∎

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

Remarks

Statement 3 is the reason the corollary is worth recording. It says that merely asking every rearrangement to converge already forces absolute convergence, so the apparently weaker demand is not weaker at all. What makes that work is the strength of 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 $\overline{R}$ : it produces not only rearrangements with prescribed sums but rearrangements with no sum.
Two of the three implications are cheap. The content is in 3 implies 1, and its only ingredient is the Riemann series theorem. The implication 1 implies 2 is Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum verbatim, and 2 implies 3 is a weakening.
Where the dividing line sits. By Absolutely convergent and conditionally convergent series, and the general starting index a convergent series is absolutely or conditionally convergent and not both, so the corollary may be read as: the conditionally convergent series are exactly the convergent series that are not unconditionally convergent. The alternating harmonic series is the standard inhabitant of that class; see FALSE: every rearrangement of a convergent series converges, and to the same sum.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources