False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: $\prod (1 + p_{k})$ converges whenever $p_{k} \to 0$

Statement

False claim: for every sequence $(p_{k})$ of reals with $p_{k} \to 0$ , the infinite product $\prod (1 + p_{k})$ converges (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors).

Equivalently, in the form it is usually met: an infinite product converges as soon as its factors tend to $1$ . That the factors tend to $1$ is necessary for convergence, and it is not sufficient. What decides the matter for nonnegative $p_{k}$ is For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent: $\prod (1 + p_{k})$ converges if and only if $\sum p_{k}$ converges.

The witness is $p_{k} : = 1 / ι (k + 1)$ , with $ι (k + 1)$ the canonical natural (Canonical naturals are positive and strictly increasing). Then $p_{k} \to 0$ , while $\sum_{k} p_{k}$ is the harmonic series $\sum_{k \geq 1} 1 / k$ , which diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ); so the product diverges, its partial products satisfying $\prod_{k < n} (1 + p_{k}) \geq 1 + \sum_{k < n} p_{k}$ and hence diverging to $+ \infty$ .

Facts & Assumptions

Given: The sequence $p_{k} : = 1 / ι (k + 1)$ , its partial sums $S_{n} = \sum_{k < n} p_{k}$ and the partial products $Π_{n} = \prod_{k < n} (1 + p_{k})$ .

[A1]

The refuted claim: if $p_{k} \to 0$ then $\prod (1 + p_{k})$ converges.

[L1]

The canonical naturals $ι (n)$ are positive for $n \geq 1$ and strictly increasing; if $0 < u < v$ then $0 < 1 / v < 1 / u$ ; 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 $1 / n < ε$ ).

[L2]

$\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ , with $ι (k)^{1} = ι (k)$ ; 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 $p > 1$ , Rational powers $a^{r}$ of a positive base, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Integer powers $a^{m}$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L3]

For nonnegative $p_{k}$ : $\prod (1 + p_{k})$ converges if and only if $\sum p_{k}$ converges, and $1 + S_{n} \leq Π_{n}$ for every $n$ (For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent).

[L4]

Convergence of an infinite product, and divergence when no tail has partial products with a nonzero limit (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors).

[L5]

For a series of nonnegative terms whose partial sums are unbounded above, those partial sums diverge to $+ \infty$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Divergence to $+ \infty$ and to $- \infty$ , Laws of finite sums and finite products).

Refutation

technique · direct

1.1

Each $p_{k} = 1 / ι (k + 1)$ is positive, and $(p_{k})$ converges to $0$ : given a rational $ε > 0$ , an $n \geq 1$ with $1 / ι (n) < ε$ satisfies $∣ p_{k} ∣ = p_{k} \leq 1 / ι (n) < ε$ for every $k \geq n$ .

givenL1

1.2

The series $\sum_{k} p_{k} = \sum_{k} 1 / ι (k + 1)$ is the $p$ -series $\sum_{k \geq 1} 1 / k$ at $p = 1$ , which diverges.

givenL2

2.1

Since the $p_{k}$ are nonnegative and $\sum p_{k}$ diverges, $\prod (1 + p_{k})$ diverges by the criterion.

step 1.1step 1.2L3

2.2

Concretely, the partial sums $S_{n}$ of the nonnegative divergent series $\sum p_{k}$ are unbounded above, so $S_{n} \to + \infty$ ; and $Π_{n} \geq 1 + S_{n}$ , so the partial products are unbounded and no tail of the product has partial products with a nonzero limit.

step 1.2L3L4L5

3.1

So $(p_{k})$ tends to $0$ while $\prod (1 + p_{k})$ diverges, and the claim [A1] is false.

step 1.1step 2.1A1

4.1

What is true is the criterion [L3]: for nonnegative terms, convergence of the product is equivalent to convergence of $\sum p_{k}$ , a strictly stronger condition than $p_{k} \to 0$ .

step 3.1A1L3∎

Remarks

The analogy with series is exact, and that is the point. For series, $a_{k} \to 0$ is necessary and not sufficient for convergence, and the harmonic series is the standard witness (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ). For products, $1 + p_{k} \to 1$ is necessary and not sufficient, and the same harmonic series is the standard witness, transported through $\prod (1 + p_{k}) \geq 1 + \sum p_{k}$ .
The failure here is unbounded growth, not oscillation. The partial products increase without limit. The other way a product can fail, with partial products tending to $0$ although the factors tend to $1$ and the series of the $p_{k}$ converges, needs the signs to alternate, and is exhibited on the companion examples page.
Nothing here needs a logarithm. The single inequality $Π_{n} \geq 1 + S_{n}$ of For $p_{k} \geq 0$ the product $\prod (1 + p_{k})$ converges iff $\sum p_{k}$ converges, with $1 + \sum_{k < n} p_{k} \leq \prod_{k < n} (1 + p_{k}) \leq 1 / (1 - \sum_{k < n} p_{k})$ when $\sum_{k < n} p_{k} < 1$ ; for $0 \leq p_{k} < 1$ the product $\prod (1 - p_{k})$ converges iff $\sum p_{k}$ converges and its partial products tend to $0$ otherwise; and $\sum ∣ p_{k} ∣$ convergent implies $\prod (1 + p_{k})$ convergent does all the work, and it is an induction on finite products.

Depends on

Used by

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Dependency tree · two levels

61 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

Infinite product (Wikipedia) (standard reference, not scraped)
Harmonic series (mathematics) (Wikipedia) (standard reference, not scraped)
Thomson, Bruckner, and Bruckner, Elementary Real Analysis (standard reference, not scraped)

FALSE: ∏(1+pk)∏(1+pk​) converges whenever pk→0pk​→0