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 convergent series converges absolutely

Statement

False claim: for every sequence (ak) of reals, if ∑ak converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then ∑ak converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index).

What is true is the converse, If ∑∣ak∣ converges then ∑ak converges: absolute convergence implies convergence. The claim above reverses it, and the reversal fails at the standard witness, the alternating harmonic series.

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), usually written εj=(−1)j, and put

aj  :=  εjι(j+1)(j∈N),

with ι(j+1) the canonical natural (Canonical naturals are positive and strictly increasing). Then ∑aj converges while ∑∣aj∣ is the harmonic series, which diverges. So the two notions really are different, and "conditionally convergent" is not an empty class.

Facts & Assumptions

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

[A1]

The refuted claim: every convergent series of reals converges absolutely.

[L2]

The canonical naturals ι(n) are positive for n≥1 and strictly increasing in n (Canonical naturals are positive and strictly increasing).

[L4]

For every real ε>0 there is a natural n≥1 with 1/ι(n)<ε (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε).

[L6]

∑k≥11/kp converges if and only if p>1, where kp=ι(k)p; at p=1 the rational power is the element itself, ι(k)1=ι(k) (For rational p>0, ∑1/kp converges iff p>1, Rational powers ar of a positive base, Existence and uniqueness of n-th roots: a unique a1/n≥0 with (a1/n)n=a, Integer powers am).

[L7]

The series ∑k≥1xk is by definition the series of the sequence j↦xj+1 (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L8]

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

[L9]

Absolute convergence means convergence of ∑∣aj∣; conditional convergence means convergence of ∑aj without it (Absolutely convergent and conditionally convergent series, and the general starting index).

[L10]

Refutation

technique · direct
1.1

Each bj=1/ι(j+1) is a positive real, ι(j+1) being a positive canonical natural.

givenL2
1.2

The sequence (bj) is nonincreasing: ι(j+1)<ι(j+2), so 1/ι(j+2)<1/ι(j+1).

L2L3
2.1

The sequence (bj) converges to 0: given a rational ε>0, fix a natural n≥1 with 1/ι(n)<ε; then for every j≥n one has ι(j+1)≥ι(n)>0, hence ∣bj∣=bj≤1/ι(n)<ε.

step 1.1L2L3L4
2.2

For every j, ∣aj∣=∣εj∣ ∣bj∣=bj=1/ι(j+1).

step 1.1L1L8
3.1

By the alternating series test, ∑aj=∑εjbj converges.

step 1.2step 2.1L5
3.2

The series ∑j1/ι(j+1) is, by the definition of a series from a general starting index, exactly the series ∑k≥11/k, that is the p-series at p=1.

step 2.2L6L7
4.1

The p-series at p=1 diverges, since 1>1 is false; so ∑∣aj∣ diverges.

step 3.2L6
5.1

Thus ∑aj converges while ∑∣aj∣ does not, so ∑aj converges conditionally and not absolutely, and the claim [A1] fails for this series.

step 3.1step 4.1A1L9
6.1

The claim is therefore false. What survives of it is only the converse implication, that an absolutely convergent series converges.

step 5.1A1L10∎

Remarks

Depends on

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Sources