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

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$\prod_{j \ge 0} \bigl(1 - 1/(j+2)\bigr)$ has partial products $1/(n+1)$ , which tend to $0$ , so the product does not converge in the sense used here

Example

Put $p_j := 1/\iota(j+2)$ , so $0 < p_j \le 1/2 < 1$ , and consider

$\prod_{j \ge 0}\Bigl(1 - \frac{1}{j+2}\Bigr) \;=\; \prod_{j \ge 0}\frac{\iota(j+1)}{\iota(j+2)} \;=\; \frac12 \cdot \frac23 \cdot \frac34 \cdots$

Its partial products telescope:

$\prod_{j<n}\Bigl(1 - \frac{1}{\iota(j+2)}\Bigr) \;=\; \frac{1}{\iota(n+1)} \qquad (n \in \mathbb{N}),$

so they tend to $0$ . Every factor is nonzero, and yet the product does not converge in the sense of Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors, because no tail of it has partial products with a nonzero limit.

This is the example the definition of a convergent infinite product is written to exclude, and Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors names it for that purpose. Were a limit of $0$ admitted, this product would "converge to $0$ " with no factor equal to $0$ , and a convergent product could no longer be divided by.

The behaviour is also exactly what For $p_k \ge 0$ the product $\prod (1 + p_k)$ converges iff $\sum p_k$ converges, with $1 + \sum_{k<n} p_k \le \prod_{k<n}(1+p_k) \le 1/\bigl(1 - \sum_{k<n} p_k\bigr)$ when $\sum_{k<n} p_k < 1$ ; for $0 \le p_k < 1$ 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 predicts: $\sum_j p_j$ is a tail of the harmonic series and diverges, so the partial products of $\prod(1-p_j)$ tend to $0$ .

Facts & Assumptions

Given: The sequence $p_j = 1/\iota(j+2)$ and the partial products $Q_n = \prod_{j<n}(1 - p_j)$ .

[L1]

Finite products: $\prod_{j<0}x_j = 1$ and $\prod_{j<n+1}x_j = \bigl(\prod_{j<n}x_j\bigr)x_n$ ; splitting at an intermediate index; a finite product of positive factors is positive (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

The canonical naturals are positive for $n \ge 1$ , strictly increasing, and $\iota(m+n) = \iota(m)+\iota(n)$ ; 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 $1/n < \varepsilon$ ).

[L3]

The principle of induction on $\mathbb{N}$ (The principle of mathematical induction).

[L4]

Convergence of an infinite product: some tail must have nonvanishing factors and partial products with a nonzero limit (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors, Limits and Cauchy sequences of reals).

[L5]

$\sum_{k\ge1}1/k^{p}$ converges if and only if $p > 1$ , with $\iota(k)^{1} = \iota(k)$ ; a series converges if and only if each of its tail series converges (For rational $p > 0$ , $\sum 1/k^p$ converges iff $p > 1$ , A series converges iff each of its tail series converges, and the sum splits as $s_N$ plus the $N$ -th tail, Rational powers $a^r$ of a positive base, Existence and uniqueness of $n$ -th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ , Integer powers $a^m$ , Series, partial sums, convergence and the sum, divergence, and the tail series).

[L6]

For $0 \le p_j < 1$ with $\sum p_j$ divergent, the partial products of $\prod(1-p_j)$ tend to $0$ (For $p_k \ge 0$ the product $\prod (1 + p_k)$ converges iff $\sum p_k$ converges, with $1 + \sum_{k<n} p_k \le \prod_{k<n}(1+p_k) \le 1/\bigl(1 - \sum_{k<n} p_k\bigr)$ when $\sum_{k<n} p_k < 1$ ; for $0 \le p_k < 1$ 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).

Verification

technique · direct

1.1

For every $j$ , $\iota(j+2) \ge \iota(2) = 2 > 0$ , so $0 < p_j \le 1/2 < 1$ and $1 - p_j = \iota(j+1)/\iota(j+2) > 0$ .

givenL2

2.1

An induction gives $Q_n = 1/\iota(n+1)$ for every $n$ : at $n = 0$ the empty product is $1 = 1/\iota(1)$ ; and $Q_{n+1} = Q_n(1 - p_n) = \dfrac{1}{\iota(n+1)}\cdot\dfrac{\iota(n+1)}{\iota(n+2)} = \dfrac{1}{\iota(n+2)}$ .

step 1.1L1L2L3

2.2

The same conclusion follows from the general criterion: $\sum_j p_j = \sum_j 1/\iota(j+2)$ is the first tail series of $\sum_j 1/\iota(j+1)$ , which is the harmonic series and diverges, so $\sum_j p_j$ diverges and the partial products of $\prod(1-p_j)$ tend to $0$ .

step 1.1L5L6

3.1

Hence $Q_n \to 0$ : given a rational $\varepsilon > 0$ , an $n_0 \ge 1$ with $1/\iota(n_0) < \varepsilon$ gives $0 < Q_n = 1/\iota(n+1) \le 1/\iota(n_0) < \varepsilon$ for every $n \ge n_0$ .

step 2.1L2

4.1

For every $N$ the $N$ -th tail products satisfy $\prod_{j=N}^{N+n-1}(1-p_j) = Q_{N+n}/Q_N$ , the finite product $Q_N$ being positive; so they also tend to $0$ as $n$ grows, $Q_N$ being a fixed nonzero real.

step 1.1step 2.1step 3.1L1

5.1

Therefore no tail of the product has partial products with a nonzero limit, and $\prod_j (1-p_j)$ does not converge, although every one of its factors is nonzero.

step 4.1L4

6.1

So the partial products are $1/\iota(n+1)$ , they tend to $0$ , and the product diverges in the sense of Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors.

step 2.1step 5.1step 2.2∎

Remarks

The telescoping is the reason the answer is exactly $1/(n+1)$ . Each factor is $\iota(j+1)/\iota(j+2)$ , so consecutive numerators and denominators cancel and only the first numerator $\iota(1) = 1$ and the last denominator $\iota(n+1)$ survive. Written informally, $\tfrac12\cdot\tfrac23\cdots\tfrac{n}{n+1} = \tfrac{1}{n+1}$ .
Why a zero limit is excluded from the definition. If it were admitted, this product would have value $0$ although no factor is $0$ ; and then from $\prod a_k = 0$ one could infer nothing about the factors, whereas Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors arranges that a convergent product is $0$ exactly when some factor is. The exclusion costs this one example and buys that.
The index shift is not decorative. Written as $\prod_{n \ge 1}(1 - 1/n)$ the same product begins with the factor $1 - 1/1 = 0$ ; the shift to $1 - 1/(j+2)$ is what keeps every factor nonzero, so that the failure is genuinely about the limit and not about a vanishing factor.

Depends on

Used by

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Sources

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

∏j≥0(1−1/(j+2))\prod_{j \ge 0} \bigl(1 - 1/(j+2)\bigr)∏j≥0​(1−1/(j+2)) has partial products 1/(n+1)1/(n+1)1/(n+1), which tend to 000, so the product does not converge in the sense used here