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

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Two series with $a_k \le b_k$ for all $k$ , $\sum b_k$ convergent and $\sum a_k$ divergent, when the terms may be negative

Statement refuted

Refuted claim: if $a_k \le b_k$ for every $k \in \mathbb{N}$ and $\sum b_k$ converges, then $\sum a_k$ converges.

This is If $0 \le a_k \le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$ , and divergence of $\sum a_k$ gives divergence of $\sum b_k$ with its nonnegativity hypothesis deleted, and deleting it destroys the theorem. Take

$a_k := -\frac{1}{k+1}, \qquad b_k := 0 \qquad (k \in \mathbb{N}) .$

Then $a_k < 0 = b_k$ for every $k$ ; the series $\sum b_k$ converges, with all partial sums equal to $0$ and sum $0$ ; and $\sum a_k$ diverges, being $-1$ times the harmonic series (The harmonic series $\sum 1/k$ diverges, by condensation and by Oresme block grouping, Convergent series add and scale termwise).

What exactly fails. The proof of the comparison test bounds the partial sums of $\sum a_k$ above by those of $\sum b_k$ and then reads convergence off boundedness, and that last step is available only for a nonnegative series (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum). Here the partial sums of $\sum a_k$ are indeed bounded above, by $0$ ; they are unbounded below, and the theorem's conclusion fails for exactly that reason.

Facts & Assumptions

Given: The sequences $a_k := -1/\iota(k+1)$ and $b_k := 0$ for $k \in \mathbb{N}$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Canonical naturals are positive and strictly increasing).

[L1]

The canonical naturals $\iota(k+1)$ are positive, so $1/\iota(k+1) > 0$ and $a_k < 0$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

A finite sum of zeros is zero, being the scalar multiple of any finite sum by $0$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L3]

A constant sequence converges to its value (Limits and Cauchy sequences of reals).

[L4]

The harmonic series $\sum_{k \ge 1} 1/k$ diverges, and it is the series of the sequence $j \mapsto 1/\iota(j+1)$ (The harmonic series $\sum 1/k$ diverges, by condensation and by Oresme block grouping, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L5]

For $c \ne 0$ , $\sum c\,x_k$ converges if and only if $\sum x_k$ converges (Convergent series add and scale termwise).

[L6]

For a series of nonnegative terms, convergence is equivalent to boundedness above of the partial sums (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L7]

The refuted claim: $a_k \le b_k$ for all $k$ and convergence of $\sum b_k$ imply convergence of $\sum a_k$ .

Counterexample

technique · direct

1.1

For every $k \in \mathbb{N}$ , $a_k = -1/\iota(k+1) < 0 = b_k$ , so in particular $a_k \le b_k$ .

givenL1

1.2

The partial sums of $\sum b_k$ are $\sum_{j<n} 0 = 0$ for every $n$ , a constant sequence, so $\sum b_k$ converges with sum $0$ .

givenL2L3

1.3

The sequence $(a_k)$ is $(-1)$ times the sequence $j \mapsto 1/\iota(j+1)$ , whose series is the harmonic series and diverges; since $-1 \ne 0$ , $\sum a_k$ diverges.

givenL4L5

2.1

So the hypotheses of the claim hold for this pair while its conclusion fails, and the claim is false.

step 1.1step 1.2step 1.3L7

3.1

The genuine comparison test is untouched: it requires $0 \le a_k$ from some index on, and here $a_k < 0$ at every index.

step 1.1L6∎

Remarks

The witness is as degenerate as possible on purpose. Taking $b_k = 0$ removes every question about the dominating series and isolates the single point at issue: a series bounded above by a convergent one need not converge if it is free to run away downwards. Any negative divergent series would do; this one is the shortest to verify.
One-sided boundedness is not convergence. The partial sums here are $-\sum_{k=1}^{n} 1/k$ , bounded above by $0$ and unbounded below. For a nonnegative series that situation cannot arise, which is exactly the content of A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum and the reason the sign hypothesis appears in every comparison statement on the main page.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 73 results over 23 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

Direct comparison test (Wikipedia) (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

Two series with ak≤bka_k \le b_kak​≤bk​ for all kkk, ∑bk\sum b_k∑bk​ convergent and ∑ak\sum a_k∑ak​ divergent, when the terms may be negative