Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: there is a divergent series of positive terms that diverges more slowly than every other, hence a universal comparison test

Statement

False claim: there is a sequence (bk) of reals with bk>0 for every k∈N such that ∑bk diverges (Series, partial sums, convergence and the sum, divergence, and the tail series) and such that every sequence (ak) of reals with ak>0 for every k and ∑ak divergent satisfies

bk≤akfor all k from some index on.

Such a (bk) would be a slowest divergent series of positive terms, and it would give a universal comparison test: a positive series would diverge exactly when its terms eventually dominate those of (bk).

No such sequence exists. The refutation is direct and uses no choice: given any divergent ∑bk with positive terms, the Abel-Dini theorem (For a divergent series of positive terms with partial sums sk, the series ∑ak/sk diverges and ∑ak/sk2 converges) manufactures a divergent series of positive terms whose terms are eventually strictly smaller than the bk, so (bk) fails its own defining property.

Facts & Assumptions

Given: An arbitrary sequence (bk) of reals with bk>0 for every k and ∑bk divergent; its inclusive partial sums Bn=∑k=0nbk (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L2]

Abel-Dini: if (ak) has positive terms and ∑ak diverges, then with Sn=∑k=0nak the series ∑nan/Sn diverges (For a divergent series of positive terms with partial sums sk, the series ∑ak/sk diverges and ∑ak/sk2 converges).

[L3]

Order and reciprocals: for x>0 and y>1 one has 0<x/y<x (Inverses of positives are positive, and reciprocation reverses order); and a sum of positive terms is positive (Laws of finite sums and finite products).

[L4]

The refuted claim: some divergent series of positive terms is eventually dominated by every divergent series of positive terms.

Refutation

technique · direct
1.1

Let (bk) be any sequence of positive reals with ∑bk divergent, and put Bn=∑k=0nbk; every Bn is positive, being a sum of positive terms.

givenL3
1.2

Since ∑bk diverges and its terms are nonnegative, its exclusive partial sums sn=∑k<nbk diverge to +∞; and Bn=sn+1, so Bn→+∞ as well, any index bound for (sn) serving for (Bn).

givenL1algebra
2.1

Define cn:=bn/Bn for n∈N. Each cn is positive, and by Abel-Dini applied to (bk) the series ∑cn diverges.

step 1.1L2L3
2.2

Since Bn→+∞ there is N∈N with Bn>1 for every n≥N; for such n, cn=bn/Bn<bn.

step 1.2L3choose
3.1

So (cn) is a sequence of positive reals with ∑cn divergent, and there is no index from which bn≤cn holds onwards: given any K, at every index n that is at least both K and N one has cn<bn.

step 2.1step 2.2
4.1

Therefore the sequence (bk) does not have the property demanded of it, and since (bk) was an arbitrary divergent series of positive terms, no such sequence exists and the claim is false.

step 3.1L4∎

Remarks

  • What this rules out. There is no fixed series against which comparison decides divergence for all positive series, so the direct comparison test is unavoidably a family of tests, one for each comparison series, with none of them final. The refutation is constructive in the strong sense: it does not merely show that a slowest series cannot exist, it exhibits, for each candidate, a specific divergent series that beats it.

  • The scale of tests on this page inherits the same limitation. Ratio, Raabe and Gauss are successive refinements, each deciding series the previous one cannot, and the argument above says the sequence of refinements can never terminate in a universal criterion. What Kummer's test adds is a uniform way of describing the whole family, by naming the weights; it does not escape the obstruction, since each choice of weights is still a comparison against the single series ∑1/ζk.

Depends on

Used by

Dependency tree · two levels

42 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