CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-08-29

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Conditional convergence of $\sum a_{n}$ does not force convergence of $\prod (1 + a_{n})$

Statement refuted

If the series $\sum a_{n}$ converges, then the product $\prod (1 + a_{n})$ converges.

Facts & Assumptions

Given: The sequence $a_{2 m - 1} = \frac{1}{\sqrt{m}}, a_{2 m} = - \frac{1}{\sqrt{m}} + \frac{1}{m^{2}} (m \geq 1) .$

[F1]

For real numbers $0 \leq p_{m} < 1$ , if $\sum p_{m}$ diverges then $\prod (1 - p_{m})$ tends to $0$ (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).

Counterexample

1.1givenalgebra

The paired partial sums satisfy $a_{2 m - 1} + a_{2 m} = 1 / m^{2}$ , so the series $\sum a_{n}$ converges because $\sum_{m \geq 1} 1 / m^{2}$ converges; it is not absolutely convergent because $\sum a_{2 m - 1} = \sum 1 / \sqrt{m}$ diverges.

2.1step 1.1algebra

The paired product is $(1 + a_{2 m - 1}) (1 + a_{2 m}) = (1 + \frac{1}{\sqrt{m}}) (1 - \frac{1}{\sqrt{m}} + \frac{1}{m^{2}}) = 1 - \frac{1}{m} + \frac{1}{m^{2}} + \frac{1}{m^{5 / 2}}$ . For all large $m$ this is at most $1 - \frac{1}{2 m}$ .

3.1F1step 2.1algebra∎

Since $\sum_{m \geq 1} 1 / (2 m)$ diverges, [F1] makes $\prod_{m \geq M} (1 - \frac{1}{2 m})$ tend to $0$ for every large $M$ ; step 2.1 therefore forces the product of the paired factors, and hence $\prod (1 + a_{n})$ itself, to tend to $0$ rather than to a nonzero limit. This refutes the statement.

Depends on

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

Used by

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Dependency tree · two levels

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Sources

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 Infinite products (standard reference, not scraped)

Conditional convergence of ∑an∑an​ does not force convergence of ∏(1+an)∏(1+an​)