False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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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 $(b_k)$ of reals with $b_k > 0$ for every $k \in \mathbb{N}$ such that $\sum b_k$ diverges (Series, partial sums, convergence and the sum, divergence, and the tail series) and such that every sequence $(a_k)$ of reals with $a_k > 0$ for every $k$ and $\sum a_k$ divergent satisfies

$b_k \le a_k \quad \text{for all } k \text{ from some index on.}$

Such a $(b_k)$ 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 $(b_k)$ .

No such sequence exists. The refutation is direct and uses no choice: given any divergent $\sum b_k$ with positive terms, the Abel-Dini theorem (For a divergent series of positive terms with partial sums $s_k$ , the series $\sum a_k/s_k$ diverges and $\sum a_k/s_k^2$ converges) manufactures a divergent series of positive terms whose terms are eventually strictly smaller than the $b_k$ , so $(b_k)$ fails its own defining property.

Facts & Assumptions

Given: An arbitrary sequence $(b_k)$ of reals with $b_k > 0$ for every $k$ and $\sum b_k$ divergent; its inclusive partial sums $B_n = \sum_{k=0}^{n} b_k$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L1]

For a series of nonnegative terms: the partial sums are nondecreasing, and if the series diverges their range is not bounded above and they diverge to $+\infty$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, A nondecreasing sequence that is not bounded above diverges to $+\infty$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Divergence to $+\infty$ and to $-\infty$ ).

[L2]

Abel-Dini: if $(a_k)$ has positive terms and $\sum a_k$ diverges, then with $S_n = \sum_{k=0}^{n} a_k$ the series $\sum_n a_n/S_n$ diverges (For a divergent series of positive terms with partial sums $s_k$ , the series $\sum a_k/s_k$ diverges and $\sum a_k/s_k^2$ 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 $(b_k)$ be any sequence of positive reals with $\sum b_k$ divergent, and put $B_n = \sum_{k=0}^{n} b_k$ ; every $B_n$ is positive, being a sum of positive terms.

givenL3

1.2

Since $\sum b_k$ diverges and its terms are nonnegative, its exclusive partial sums $s_n = \sum_{k<n} b_k$ diverge to $+\infty$ ; and $B_n = s_{n+1}$ , so $B_n \to +\infty$ as well, any index bound for $(s_n)$ serving for $(B_n)$ .

givenL1algebra

2.1

Define $c_n := b_n / B_n$ for $n \in \mathbb{N}$ . Each $c_n$ is positive, and by Abel-Dini applied to $(b_k)$ the series $\sum c_n$ diverges.

step 1.1L2L3

2.2

Since $B_n \to +\infty$ there is $N \in \mathbb{N}$ with $B_n > 1$ for every $n \ge N$ ; for such $n$ , $c_n = b_n/B_n < b_n$ .

step 1.2L3choose

3.1

So $(c_n)$ is a sequence of positive reals with $\sum c_n$ divergent, and there is no index from which $b_n \le c_n$ holds onwards: given any $K$ , at every index $n$ that is at least both $K$ and $N$ one has $c_n < b_n$ .

step 2.1step 2.2

4.1

Therefore the sequence $(b_k)$ does not have the property demanded of it, and since $(b_k)$ 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 $\sum 1/\zeta_k$ .

Depends on

Used by

How the nonnegative tests are ordered by strength, and which of them this page cannot state without the logarithm Remark

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 83 results over 25 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Divergent series (Wikipedia) (standard reference, not scraped)
K. Knopp, Theory and Application of Infinite Series, Ch. IX (standard reference, not scraped)
Abel-Dini-Pringsheim theorem (Wikipedia) (standard reference, not scraped)