TheoremStatement: AI-adaptedProof: AI-adaptedverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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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

Statement

Write $S_{n} : = \sum_{k < n} p_{k}$ and $Π_{n} : = \prod_{k < n} (1 + p_{k})$ , $Q_{n} : = \prod_{k < n} (1 - p_{k})$ (Finite sums and finite products, by recursion, Series, partial sums, convergence and the sum, divergence, and the tail series, Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors).

Elementary inequalities. Let $p_{k} \geq 0$ for every $k$ . Then for every $n \in N$ : $1 + S_{n} \leq Π_{n}, and Π_{n} \leq \frac{1}{1 - S_{n}} whenever S_{n} < 1;$ and if in addition $p_{k} \leq 1$ for every $k$ , then $1 - S_{n} \leq Q_{n} and Q_{n} Π_{n} \leq 1, hence Q_{n} \leq \frac{1}{1 + S_{n}} .$
The nonnegative criterion. Let $p_{k} \geq 0$ for every $k$ . Then $\prod (1 + p_{k})$ converges if and only if $\sum p_{k}$ converges.
The $(1 - p_{k})$ form. Let $0 \leq p_{k} < 1$ for every $k$ . Then $\prod (1 - p_{k})$ converges if and only if $\sum p_{k}$ converges; and if $\sum p_{k}$ diverges then $Q_{n} \to 0$ , so that no tail of the product has partial products with a nonzero limit.
Absolute convergence. Let $(p_{k})$ be an arbitrary sequence of reals with $\sum ∣ p_{k} ∣$ convergent. Then $\prod (1 + p_{k})$ converges.

No logarithm occurs anywhere. The exponential and the logarithm, through which these criteria are usually derived, are later in the reading order; every inequality above is an induction on finite products. The refinement that decides $\prod (1 + p_{k})$ for signed $p_{k}$ with $\sum p_{k}$ convergent, in terms of the convergence of $\sum p_{k}^{2}$ , does need the logarithm and is not stated here; see Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for.

Facts & Assumptions

Given: A sequence $(p_{k})$ of reals, with $S_{n} = \sum_{k < n} p_{k}$ , $Π_{n} = \prod_{k < n} (1 + p_{k})$ and $Q_{n} = \prod_{k < n} (1 - p_{k})$ .

[L1]

Finite sums and products: $\sum_{k < 0} x_{k} = 0$ , $\prod_{k < 0} x_{k} = 1$ , $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ , $\prod_{k < n + 1} x_{k} = (\prod_{k < n} x_{k}) x_{n}$ , splitting at an intermediate index, and $\prod_{k < n} (x_{k} y_{k}) = (\prod_{k < n} x_{k}) (\prod_{k < n} y_{k})$ ; a finite product of nonnegative factors is nonnegative and of positive factors is positive (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

The principle of induction on $N$ (The principle of mathematical induction).

[L3]

For a series of nonnegative terms: convergence is equivalent to the range of the partial sums being bounded above, the sum is then the supremum and every partial sum is at most the sum, and if the range is unbounded the 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 $- \infty$ ).

[L4]

A series converges if and only if some tail series converges, and then the sum equals the initial partial sum plus the tail sum (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail).

[L5]

A nondecreasing sequence bounded above converges, and a nonincreasing sequence bounded below converges; a monotone sequence converges if and only if it is bounded (A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum, A monotone sequence converges if and only if it is bounded, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L6]

Order and inverses: $0 < a < b$ implies $0 < 1 / b < 1 / a$ , and $a > 0$ implies $1 / a > 0$ (Inverses of positives are positive, and reciprocation reverses order).

[L7]

Absolute value: $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , $∣ x ∣ \geq 0$ , $- ∣ x ∣ \leq x \leq ∣ x ∣$ , and $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ (Basic properties of the absolute value).

[L8]

Algebra of limits, and limits preserve non-strict inequalities holding eventually (Algebra of limits: sums, scalar multiples, products and quotients, Limits preserve non-strict inequalities, Limits and Cauchy sequences of reals).

[L9]

The squeeze theorem (The squeeze theorem).

[L10]

Every Cauchy sequence of reals converges (The reals are complete, Limits and Cauchy sequences of reals).

[L11]

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

Proof

technique · direct

1.1

Assume $p_{k} \geq 0$ for every $k$ . An induction gives $1 + S_{n} \leq Π_{n}$ : at $n = 0$ both sides are $1$ ; and if $1 + S_{n} \leq Π_{n}$ then, since $1 + S_{n} \geq 1 > 0$ and $1 + p_{n} \geq 1 > 0$ , $Π_{n + 1} = Π_{n} (1 + p_{n}) \geq (1 + S_{n}) (1 + p_{n}) = 1 + S_{n} + p_{n} + S_{n} p_{n} \geq 1 + S_{n + 1}$ .

givenL1L2

1.2

Assume further $p_{k} \leq 1$ for every $k$ . An induction gives $1 - S_{n} \leq Q_{n}$ : at $n = 0$ both sides are $1$ ; and if $1 - S_{n} \leq Q_{n}$ then, since $1 - p_{n} \geq 0$ , $Q_{n + 1} = Q_{n} (1 - p_{n}) \geq (1 - S_{n}) (1 - p_{n}) = 1 - S_{n} - p_{n} + S_{n} p_{n} \geq 1 - S_{n + 1}$ .

givenL1L2

1.3

Assume $p_{k} \geq 0$ and $\sum p_{k}$ convergent, with sum $L$ . By [L3] and [L4] the tail sums $L - S_{N}$ tend to $0$ , so fix $N$ with $\sum_{k \geq N} p_{k} < 1 / 2$ .

givenL3L4choose

1.4

Three inductions on finite products, valid for arbitrary reals $x_{k}, y_{k}, z_{k}$ : first, $∣ \prod_{k < n} x_{k} ∣ = \prod_{k < n} ∣ x_{k} ∣$ , from $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ and $∣ 1 ∣ = 1$ ; second, if $0 \leq x_{k} \leq y_{k}$ for all $k < n$ then $\prod_{k < n} x_{k} \leq \prod_{k < n} y_{k}$ , since the products are nonnegative and $\prod_{k < n + 1} x_{k} = (\prod_{k < n} x_{k}) x_{n} \leq (\prod_{k < n} y_{k}) x_{n} \leq (\prod_{k < n} y_{k}) y_{n}$ ; third, $∣ \prod_{k < n} (1 + z_{k}) - 1 ∣ \leq \prod_{k < n} (1 + ∣ z_{k} ∣) - 1$ , since at $n = 0$ both sides are $0$ and $∣ \prod_{k < n + 1} (1 + z_{k}) - 1 ∣ = ∣ (\prod_{k < n} (1 + z_{k}) - 1) (1 + z_{n}) + z_{n} ∣ \leq (\prod_{k < n} (1 + ∣ z_{k} ∣) - 1) (1 + ∣ z_{n} ∣) + ∣ z_{n} ∣ = \prod_{k < n + 1} (1 + ∣ z_{k} ∣) - 1$ .

L1L2L7

1.5

Assume $\sum ∣ p_{k} ∣$ converges, with sum $L$ , and fix $N$ with $τ_{N} : = \sum_{k \geq N} ∣ p_{k} ∣ < 1 / 2$ ; write $τ_{N + n} = \sum_{k \geq N + n} ∣ p_{k} ∣$ , so $τ_{N + n} \to 0$ and $\sum_{j < m} ∣ p_{N + j} ∣ \leq τ_{N}$ for every $m$ . For $k \geq N$ we get $∣ p_{k} ∣ \leq τ_{N} < 1 / 2$ , so $1 + p_{k} \geq 1 / 2 > 0$ and every factor from $N$ on is nonzero.

givenL3L4L7choose

2.1

An induction gives: for every $n$ with $S_{n} < 1$ , $Π_{n} (1 - S_{n}) \leq 1$ . At $n = 0$ this reads $1 \cdot 1 \leq 1$ . Suppose it holds at $n$ and $S_{n + 1} < 1$ ; then $S_{n} \leq S_{n + 1} < 1$ , so $Π_{n} \leq 1 / (1 - S_{n})$ by [L6], and $Π_{n + 1} = Π_{n} (1 + p_{n}) \leq (1 + p_{n}) / (1 - S_{n})$ . Multiplying out, $(1 + p_{n}) (1 - S_{n} - p_{n}) = 1 - S_{n} - p_{n} S_{n} - p_{n}^{2} \leq 1 - S_{n}$ , and dividing by the positive $(1 - S_{n}) (1 - S_{n + 1})$ turns this into $(1 + p_{n}) / (1 - S_{n}) \leq 1 / (1 - S_{n + 1})$ .

givenstep 1.1L1L2L6

2.2

Under the same assumption, $Q_{n} Π_{n} = \prod_{k < n} (1 - p_{k}) (1 + p_{k}) = \prod_{k < n} (1 - p_{k}^{2}) \leq 1$ , the last step by the induction: the empty product is $1$ , and multiplying a value in $[0, 1]$ by a factor $1 - p_{n}^{2} \in [0, 1]$ again gives a value in $[0, 1]$ . Since $Π_{n} \geq 1 + S_{n} \geq 1 > 0$ , dividing gives $Q_{n} \leq 1 / Π_{n} \leq 1 / (1 + S_{n})$ . This completes claim 1.

step 1.1step 1.2L1L2L6

2.3

Assume $0 \leq p_{k} < 1$ and $\sum p_{k}$ convergent. Fix $N$ with $\sum_{k \geq N} p_{k} < 1 / 2$ as in step 1.3. By step 1.2 applied to the shifted sequence, $U_{n} : = \prod_{j < n} (1 - p_{N + j}) \geq 1 - \sum_{j < n} p_{N + j} \geq 1 / 2$ for every $n$ ; and $(U_{n})$ is nonincreasing, each factor lying in $(0, 1]$ . So $(U_{n})$ converges to a limit $\geq 1 / 2 > 0$ , and every factor $1 - p_{k}$ is positive, hence nonzero; $\prod (1 - p_{k})$ converges.

step 1.2step 1.3L1L5L8L11

3.1

For the shifted sequence $j \mapsto p_{N + j}$ , whose partial sums are at most $1 / 2 < 1$ , step 2.1 gives $T_{n} : = \prod_{j < n} (1 + p_{N + j}) \leq 1 / (1 - 1 / 2) = 2$ for every $n$ , and step 1.1 gives $T_{n} \geq 1$ . The sequence $(T_{n})$ is nondecreasing, each factor being at least $1$ , so it converges to a limit $ℓ$ with $1 \leq ℓ \leq 2$ ; in particular $ℓ \neq 0$ , and every factor $1 + p_{k}$ is at least $1$ , hence nonzero. So $\prod (1 + p_{k})$ converges.

step 1.1step 2.1step 1.3L1L5L8L11

3.2

Assume instead $0 \leq p_{k} < 1$ and $\sum p_{k}$ divergent. Then $S_{n} \to + \infty$ by [L3], so given a real $ε > 0$ there is $K$ with $S_{n} > 1 / ε$ for $n \geq K$ , whence $0 < 1 / (1 + S_{n}) < ε$ ; thus $1 / (1 + S_{n}) \to 0$ . By step 2.2, $0 \leq Q_{n} \leq 1 / (1 + S_{n})$ , so $Q_{n} \to 0$ by the squeeze.

step 2.2L3L6L9

3.3

Put $T_{n} : = \prod_{j < n} (1 + p_{N + j})$ . By step 1.4 and step 2.1 applied to the nonnegative sequence $j \mapsto ∣ p_{N + j} ∣$ , $∣ T_{n} ∣ \leq \prod_{j < n} (1 + ∣ p_{N + j} ∣) \leq 1 / (1 - τ_{N}) \leq 2$ ; and by step 1.4 and step 1.2, $T_{n} \geq \prod_{j < n} (1 - ∣ p_{N + j} ∣) \geq 1 - τ_{N} \geq 1 / 2$ , each factor $1 + p_{N + j} \geq 1 - ∣ p_{N + j} ∣ \geq 0$ .

step 2.1step 1.2step 1.4step 1.5L1L7

4.1

Conversely assume $p_{k} \geq 0$ and $\prod (1 + p_{k})$ convergent, with $N$ as in [L11]. Since $Π_{n} = (\prod_{k < N} (1 + p_{k})) T_{n - N}$ for $n \geq N$ and $(T_{m})$ converges, the sequence $(Π_{n})$ converges, hence is bounded, say $Π_{n} \leq M$ for all $n$ . By step 1.1, $1 + S_{n} \leq M$ for every $n$ , so the partial sums of the nonnegative series $\sum p_{k}$ are bounded above and $\sum p_{k}$ converges. Claim 2 is step 3.1 together with this.

step 1.1step 3.1L1L3L5L11

4.2

In that situation the product diverges: for any $N$ , $Q_{N + n} = (\prod_{k < N} (1 - p_{k})) U_{n}$ with $\prod_{k < N} (1 - p_{k}) > 0$ , so $U_{n} = Q_{N + n} / \prod_{k < N} (1 - p_{k}) \to 0$ and no tail has partial products with a nonzero limit. With step 2.3 this proves claim 3.

step 2.3step 3.2L1L8L11

4.3

For $m > n$ , splitting the product gives $T_{m} = T_{n} \prod_{j = n}^{m - 1} (1 + p_{N + j})$ , so $∣ T_{m} - T_{n} ∣ = ∣ T_{n} ∣ ∣ \prod_{j = n}^{m - 1} (1 + p_{N + j}) - 1 ∣ \leq 2 (\prod_{j = n}^{m - 1} (1 + ∣ p_{N + j} ∣) - 1) \leq 2 (\frac{1}{1 - τ_{N + n}} - 1) = \frac{2 τ_{N + n}}{1 - τ_{N + n}} \leq 4 τ_{N + n}$ , using step 2.1 for the shifted sequence from $N + n$ , whose partial sums are at most $τ_{N + n} \leq τ_{N} < 1 / 2$ .

step 2.1step 1.4step 1.5step 3.3L1L6L7

5.1

Since $τ_{N + n} \to 0$ , step 4.3 makes $(T_{n})$ a Cauchy sequence, so it converges, to a limit $ℓ$ ; and $ℓ \geq 1 / 2 > 0$ by step 3.3 and [L8]. Hence $\prod (1 + p_{k})$ converges, which is claim 4.

step 1.5step 3.3step 4.3L8L10L11∎

Remarks

Why the two bounds of claim 1 are the right pair. The lower bound $1 + S_{n} \leq Π_{n}$ is the Weierstrass product inequality and forces divergence of the product when $\sum p_{k}$ diverges; the upper bound $Π_{n} \leq 1 / (1 - S_{n})$ , available once the partial sums are below $1$ , forces convergence when $\sum p_{k}$ converges. Between them they prove claim 2 with no further input, and they are exactly what a logarithm would otherwise supply.
The strict inequality $p_{k} < 1$ keeps this proof uniform, but the tail-based definition allows a slightly stronger statement. Claim 3 remains true for $0 \leq p_{k} \leq 1$ . If only finitely many $p_{k}$ equal $1$ , start the product after the last zero factor; if infinitely many do, then $\sum p_{k}$ diverges and no tail has all factors nonzero. The stated strict form avoids this finite/infinite split.
Claim 4 does not identify the value, and the converse fails. Absolute convergence of $\sum p_{k}$ gives convergence of $\prod (1 + p_{k})$ , but convergence of $\sum p_{k}$ alone does not: the companion examples page exhibits $\sum_{j \geq 0} (- 1)^{j} / \sqrt{j + 2}$ convergent while the corresponding partial products tend to $0$ . What separates the two cases is the convergence of $\sum p_{k}^{2}$ , a criterion that needs the logarithm and is deferred.
Where the Cauchy criterion enters and why nothing cheaper would do. In claim 4 the factors have no sign, so the partial products are not monotone and A nondecreasing sequence bounded above converges to the supremum of its range, and a nonincreasing sequence bounded below to the infimum is unavailable; the estimate of step 4.3 is a Cauchy estimate and is closed by completeness of $R$ .

Depends on

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65 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)
Weierstrass product inequality (Wikipedia) (standard reference, not scraped)
Thomson, Bruckner, and Bruckner, Elementary Real Analysis (standard reference, not scraped)
D. Dikranjan, Analysis 478, Chapter 6 (standard reference, not scraped)