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Absolute convergence criterion for complex infinite products

Statement

Let $(a_{n})_{n \geq 0}$ be a sequence of complex numbers. The following are equivalent:

the product $\prod_{n \geq 0} (1 + a_{n})$ is absolutely convergent, meaning that $\prod_{n \geq 0} (1 + ∣ a_{n} ∣)$ converges;
the series $\sum_{n \geq 0} ∣ a_{n} ∣$ converges.

When these conditions hold, the complex product $\prod_{n \geq 0} (1 + a_{n})$ itself converges. Its value is nonzero if and only if $1 + a_{n} \neq 0$ for every $n \geq 0$ .

Facts & Assumptions

Given: A complex sequence $(a_{n})$ .

[F1]

Absolute convergence of $\prod (1 + a_{n})$ means convergence of the real product $\prod (1 + ∣ a_{n} ∣)$ (Complex infinite-product convention extending the published real definition).

[F2]

For nonnegative reals $(p_{n})$ , the product $\prod (1 + p_{n})$ converges if and only if the series $\sum p_{n}$ converges; also, when $\sum p_{n}$ converges, every tail with sufficiently small sum has bounded partial products (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).

[F3]

An infinite product converges when some tail has nonzero factors and a nonzero tail-product limit (Infinite products: partial products, and convergence to a nonzero limit after finitely many vanishing factors).

Proof

technique · direct

1.1F1F2given

By [F1], absolute convergence of $\prod (1 + a_{n})$ is exactly convergence of the real product $\prod (1 + ∣ a_{n} ∣)$ , and [F2] makes that equivalent to convergence of $\sum ∣ a_{n} ∣$ . This proves the equivalence of claims 1 and 2.

1.2F2choosealgebra

Assume now that $\sum ∣ a_{n} ∣$ converges. By [F2], choose $N$ so that $\sum_{n \geq N} ∣ a_{n} ∣ < 1 / 2$ ; then $∣ a_{n} ∣ < 1 / 2$ for every $n \geq N$ , so $1 + a_{n} \neq 0$ on that tail.

2.1F2step 1.2algebra

For $m > n \geq N$ one has $∣ \prod_{k = n}^{m} (1 + a_{k}) - 1 ∣ \leq \prod_{k = n}^{m} (1 + ∣ a_{k} ∣) - 1$ , and the right-hand side tends to $0$ as $n, m \to \infty$ because the real tail products converge by [F2]. Hence the complex tail partial products form a Cauchy sequence, so they converge to some limit $ℓ \in C$ .

3.1F2F3step 1.2step 2.1algebra∎

For the same tail, $∣ 1 + a_{n} ∣ \geq 1 - ∣ a_{n} ∣ > 0$ , so $∣ \prod_{k = N}^{m} (1 + a_{k}) ∣ \geq \prod_{k = N}^{m} (1 - ∣ a_{k} ∣)$ for every $m \geq N$ ; by [F2], the real product $\prod_{k \geq N} (1 - ∣ a_{k} ∣)$ converges to a positive limit because $\sum_{k \geq N} ∣ a_{k} ∣$ converges and each term is in $[0, 1 / 2)$ . Therefore the complex tail partial products are bounded away from $0$ , so the limit $ℓ$ of step 2.1 is nonzero. Now [F3] makes $\prod (1 + a_{n})$ convergent with value $(\prod_{k < N} (1 + a_{k})) ℓ$ . Since $ℓ \neq 0$ , this value is nonzero exactly when the finite head has no zero factor; the tail factors are already nonzero by step 1.2.

Depends on

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Sources

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 Infinite products (standard reference, not scraped)
Matthias Weber, Complex Analysis, Ch. 3 §3.2 (standard reference, not scraped)