CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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With $a_{j} = (- 1)^{j} / \sqrt{j + 1}$ convergent and $b_{j} = (- 1)^{j}$ bounded but not monotone, $\sum a_{j} b_{j} = \sum 1 / \sqrt{j + 1}$ diverges

Statement refuted

Refuted claim: if $\sum a_{j}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) and $(b_{j})$ is bounded (Lower bound, bounded below, bounded set), then $\sum a_{j} b_{j}$ converges.

This is Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges with the word monotone deleted from its hypothesis on $(b_{j})$ . Deleting it destroys the theorem.

Let $(ε_{j})$ 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 $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ) and put

$a_{j} : = \frac{ε_{j}}{\sqrt{ι (j + 1)}}, b_{j} : = ε_{j} .$

Then $\sum a_{j}$ converges by the alternating series test, $(b_{j})$ is bounded with $∣ b_{j} ∣ = 1$ , and

$a_{j} b_{j} = \frac{ε_{j}^{2}}{\sqrt{ι (j + 1)}} = \frac{1}{\sqrt{ι (j + 1)}},$

so $\sum_{j} a_{j} b_{j}$ is $\sum_{k \geq 1} 1 / k^{1 / 2}$ , the $p$ -series at $p = 1 / 2$ , which diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

Facts & Assumptions

Given: The alternating sequence $(ε_{j})$ , the sequence $β_{j} : = 1 / \sqrt{ι (j + 1)}$ , and $a_{j} : = ε_{j} β_{j}$ , $b_{j} : = ε_{j}$ .

[L1]

The alternating sequence: $ε_{0} = 1$ , $ε_{j + 1} = - ε_{j}$ , $∣ ε_{j} ∣ = 1$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

Square roots: every $t \geq 0$ has a unique $\sqrt{t} \geq 0$ with $(\sqrt{t})^{2} = t$ , and $\sqrt{t} = t^{1 / 2}$ in the notation of rational powers (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Rational powers $a^{r}$ of a positive base).

[L3]

The canonical naturals are positive for $n \geq 1$ and strictly increasing; reciprocation reverses the order on the positives; and for every real $ε > 0$ there is $n \geq 1$ with $1 / ι (n) < ε$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L4]

The alternating series test (The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Limits and Cauchy sequences of reals).

[L5]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ ; and $\sum_{k \geq 1} x_{k}$ is the series of $j \mapsto x_{j + 1}$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L6]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , $∣ x ∣ \geq 0$ , and $x^{2} = ∣ x ∣^{2}$ (Basic properties of the absolute value).

[L7]

Abel's test, whose hypothesis on the second factor is that it be monotone and bounded (Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences, Lower bound, bounded below, bounded set).

Counterexample

technique · direct

1.1

Square roots are strictly increasing on the nonnegative reals: if $0 \leq u < v$ and $\sqrt{u} \geq \sqrt{v}$ then $u = (\sqrt{u})^{2} \geq (\sqrt{v})^{2} = v$ , which is false.

1.2

The sequence $(b_{j}) = (ε_{j})$ is bounded, $∣ b_{j} ∣ = 1$ for every $j$ .

L1L6

1.3

It is not monotone: $b_{0} = 1 > b_{1} = - 1$ , so it is not nondecreasing, and $b_{1} = - 1 < b_{2} = 1$ , so it is not nonincreasing.

2.1

Each $β_{j} = 1 / \sqrt{ι (j + 1)}$ is positive and $(β_{j})$ is nonincreasing, since $0 < ι (j + 1) < ι (j + 2)$ gives $0 < \sqrt{ι (j + 1)} < \sqrt{ι (j + 2)}$ .

step 1.1L2L3

2.2

$(β_{j})$ converges to $0$ : given a rational $ε > 0$ , fix $n \geq 1$ with $1 / ι (n) < ε^{2}$ ; for $j \geq n$ one has $ι (j + 1) \geq ι (n) > (1 / ε)^{2}$ , so $\sqrt{ι (j + 1)} > 1 / ε$ and $β_{j} < ε$ .

step 1.1L2L3

3.1

By the alternating series test $\sum_{j} a_{j} = \sum_{j} ε_{j} β_{j}$ converges.

step 2.1step 2.2L4

3.2

For every $j$ , $a_{j} b_{j} = ε_{j}^{2} β_{j} = ∣ ε_{j} ∣^{2} β_{j} = β_{j} = 1 / \sqrt{ι (j + 1)}$ .

step 2.1L1L6

4.1

The series $\sum_{j} β_{j}$ is $\sum_{k \geq 1} 1 / \sqrt{k} = \sum_{k \geq 1} 1 / k^{1 / 2}$ , the $p$ -series at $p = 1 / 2$ ; since $1 / 2 > 1$ is false, it diverges.

step 3.2L2L5

5.1

So $\sum a_{j}$ converges and $(b_{j})$ is bounded, while $\sum a_{j} b_{j}$ diverges: the refuted claim fails, and the hypothesis of [L7] that is missing is precisely monotonicity of $(b_{j})$ .

step 3.1step 1.2step 1.3step 4.1L7∎

Remarks

The failure is not a matter of size. The factor $(b_{j})$ has absolute value exactly $1$ at every index, so it neither grows nor shrinks; what it does is cancel the alternation of $(a_{j})$ , and the alternation was the only reason $\sum a_{j}$ converged. A monotone factor cannot do that, which is the content of Abel's test: if $\sum a_{k}$ converges and $(b_{k})$ is monotone and bounded then $\sum a_{k} b_{k}$ converges.
The exponent $1 / 2$ is chosen so that both halves work. A faster decay, such as $1 / ι (j + 1)$ , would still give a divergent product series, while a faster one still, such as $1 / ι (j + 1)^{2}$ , would give an absolutely convergent $\sum a_{j}$ , for which no bounded factor can destroy convergence. The witness has to sit in the conditionally convergent range, and $p = 1 / 2$ does.
The same series appears again on this page. The Cauchy product of $\sum_{k \geq 0} (- 1)^{k} / \sqrt{k + 1}$ with itself has $∣ c_{n} ∣ \geq 1$ for every $n$ , so it diverges uses $\sum ε_{k} / \sqrt{ι (k + 1)}$ as both factors of a Cauchy product, for the same underlying reason: its terms are just barely small enough to converge with signs and not without.

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Sources

Abel's test (Wikipedia) (standard reference, not scraped)
Harmonic series (mathematics) (Wikipedia) (standard reference, not scraped)

With aj=(−1)j/j+1aj​=(−1)j/j+1​ convergent and bj=(−1)jbj​=(−1)j bounded but not monotone, ∑ajbj=∑1/j+1∑aj​bj​=∑1/j+1​ diverges