TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Under square summability, the signed product of (1+p_n) converges iff the series of p_n converges

Statement

Let $(p_{n})$ be a real sequence such that $\sum_{n} p_{n}^{2}$ converges. Then $\prod_{n} (1 + p_{n}) converges ⟺ \sum_{n} p_{n} converges .$ The product uses the tail convention of Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors, so finitely many zero factors are allowed.

Facts & Assumptions

Given: A real sequence $(p_{n})$ with $\sum_{n} p_{n}^{2}$ convergent.

[L1]

A convergent series has terms tending to zero (If a series converges then its terms tend to $0$ ).

[L2]

$\log (1 + t) = t + \sum_{k \geq 2} (- 1)^{k + 1} t^{k} / k$ for $∣ t ∣ < 1$ ([The power series for log(1+x) on (-1,1], including the Abel endpoint](/item/thm-log-one-plus-x-power-series)).

[L3]

Direct comparison and absolute convergence give convergence of a series dominated by a convergent nonnegative series (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ , If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

[L4]

Convergent series are closed under termwise addition and subtraction, and deleting a finite initial segment preserves convergence (Convergent series add and scale termwise, A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L5]

$\exp (u + v) = \exp (u) \exp (v)$ , while $\exp$ and $\log$ are continuous inverse functions on their stated domains (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ , The exponential function is strictly increasing, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L6]

An infinite product converges when a tail of nonzero factors has nonzero limiting tail products; initial factors may be arbitrary (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors).

Proof

technique · direct

1.1

By [L1], choose $N$ such that $∣ p_{n} ∣ \leq 1 / 2$ for $n \geq N$ ; then $1 + p_{n} \geq 1 / 2 > 0$ on that tail.

L1choose

1.2

For $∣ t ∣ \leq 1 / 2$ , the tail of the series in [L2] has absolute value at most $\sum_{k \geq 2} ∣ t ∣^{k} \leq 2 t^{2}$ , hence $∣ \log (1 + t) - t ∣ \leq 2 t^{2}$ .

L2algebra

1.3

Conversely, a nonzero limit of those positive tail products has a logarithm; continuity of $\log$ and the same finite-product identity make the tail logarithm partial sums converge.

L5L6

2.1

Applying step 1.2 to $p_{n}$ and using [L3] shows that $\sum_{n \geq N} (\log (1 + p_{n}) - p_{n})$ converges absolutely.

step 1.1step 1.2L3

3.1

By [L4], $\sum_{n} p_{n}$ converges if and only if $\sum_{n \geq N} \log (1 + p_{n})$ converges.

step 2.1L4

4.1

The $m$ -th tail product equals $\exp (\sum_{n = N}^{N + m - 1} \log (1 + p_{n}))$ by repeated use of [L5], so convergence of the logarithm series gives a nonzero tail-product limit.

step 3.1L5L6

5.1

Steps 3.1, 4.1, and 1.3 prove both directions, and [L6] makes finite initial zero factors harmless.

step 3.1step 4.1step 1.3L6∎

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Sources

J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)