CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $(\varepsilon_j)$ be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ ) and put

$a_j := \frac{\varepsilon_j}{\sqrt{\iota(j+1)}}, \qquad b_j := \varepsilon_j .$

Then $\sum a_j$ converges by the alternating series test, $(b_j)$ is bounded with $|b_j| = 1$ , and

$a_j b_j \;=\; \frac{\varepsilon_j^{\,2}}{\sqrt{\iota(j+1)}} \;=\; \frac{1}{\sqrt{\iota(j+1)}},$

so $\sum_j a_j b_j$ is $\sum_{k \ge 1} 1/k^{1/2}$ , 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 $(\varepsilon_j)$ , the sequence $\beta_j := 1/\sqrt{\iota(j+1)}$ , and $a_j := \varepsilon_j \beta_j$ , $b_j := \varepsilon_j$ .

[L1]

The alternating sequence: $\varepsilon_0 = 1$ , $\varepsilon_{j+1} = -\varepsilon_j$ , $|\varepsilon_j| = 1$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ ).

[L2]

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

[L3]

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

[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]

$\sum_{k\ge1}1/k^{p}$ converges if and only if $p > 1$ ; and $\sum_{k\ge1}x_k$ 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: $|xy| = |x|\,|y|$ , $|x| \ge 0$ , 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 $0 \le u < v$ and $\sqrt u \ge \sqrt v$ then $u = (\sqrt u)^{2} \ge (\sqrt v)^{2} = v$ , which is false.

1.2

The sequence $(b_j) = (\varepsilon_j)$ 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 $\beta_j = 1/\sqrt{\iota(j+1)}$ is positive and $(\beta_j)$ is nonincreasing, since $0 < \iota(j+1) < \iota(j+2)$ gives $0 < \sqrt{\iota(j+1)} < \sqrt{\iota(j+2)}$ .

step 1.1L2L3

2.2

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

step 1.1L2L3

3.1

By the alternating series test $\sum_j a_j = \sum_j \varepsilon_j \beta_j$ converges.

step 2.1step 2.2L4

3.2

For every $j$ , $a_j b_j = \varepsilon_j^{\,2}\beta_j = |\varepsilon_j|^{2}\beta_j = \beta_j = 1/\sqrt{\iota(j+1)}$ .

step 2.1L1L6

4.1

The series $\sum_j \beta_j$ is $\sum_{k \ge 1} 1/\sqrt{k} = \sum_{k\ge1} 1/k^{1/2}$ , 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 $1/\iota(j+1)$ , would still give a divergent product series, while a faster one still, such as $1/\iota(j+1)^{2}$ , 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 $\sum_{k \ge 0} (-1)^{k}/\sqrt{k+1}$ with itself has $|c_n| \ge 1$ for every $n$ , so it diverges uses $\sum \varepsilon_k/\sqrt{\iota(k+1)}$ 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.

Depends on

Used by

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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+1a_j = (-1)^{j}/\sqrt{j+1}aj​=(−1)j/j+1​ convergent and bj=(−1)jb_j = (-1)^{j}bj​=(−1)j bounded but not monotone, ∑ajbj=∑1/j+1\sum a_j b_j = \sum 1/\sqrt{j+1}∑aj​bj​=∑1/j+1​ diverges