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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

If $\sum a_k$ and $\sum b_k$ both converge absolutely then their Cauchy product converges absolutely, with sum $AB$

Statement

Let $(a_k)$ and $(b_k)$ be sequences of reals whose series both converge absolutely (Absolutely convergent and conditionally convergent series, and the general starting index), with sums $A$ and $B$ , and let $(c_n)$ be their Cauchy product (The Cauchy product of two series: $c_n = \sum_{k=0}^{n} a_k b_{n-k}$ ). Then $\sum c_n$ converges absolutely, and

$\sum_{n=0}^{\infty} c_n \;=\; A\,B .$

Moreover $\sum_{n=0}^{\infty} |c_n| \le \bigl(\sum_{k=0}^{\infty}|a_k|\bigr) \bigl(\sum_{k=0}^{\infty}|b_k|\bigr)$ .

Combined with Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum this says that within the absolutely convergent series the product behaves exactly as one would want: it converges, its sum is the product of the sums, and neither factor's order nor the product's order matters.

Facts & Assumptions

Given: Sequences $(a_k)$ and $(b_k)$ with $\sum |a_k|$ and $\sum |b_k|$ convergent, sums $L_a$ and $L_b$ respectively, partial sums $P_N = \sum_{k<N}|a_k|$ and $Q_m = \sum_{j<m}|b_j|$ , and the Cauchy product $c_n = \sum_{k=0}^{n} a_k b_{n-k}$ (The Cauchy product of two series: $c_n = \sum_{k=0}^{n} a_k b_{n-k}$ ).

[L1]

The finite identity of Mertens' theorem: if $\sum a_k$ converges absolutely to $A$ and $\sum b_k$ converges to $B$ , their Cauchy product converges to $AB$ , claim 1: for arbitrary sequences $(x_k)$ , $(y_k)$ with partial sums $Y_m = \sum_{j<m} y_j$ and Cauchy product $(z_n)$ , one has $\sum_{n<N} z_n = \sum_{i<N} x_i\, Y_{N-i}$ for every $N$ .

[L2]

Mertens' theorem, claim 2 of Mertens' theorem: if $\sum a_k$ converges absolutely to $A$ and $\sum b_k$ converges to $B$ , their Cauchy product converges to $AB$ : if $\sum x_k$ converges absolutely and $\sum y_k$ converges, their Cauchy product converges to the product of the sums.

[L3]

$\bigl|\sum_{k<n} x_k\bigr| \le \sum_{k<n}|x_k|$ (Triangle inequality for finite sums).

[L4]

Absolute value: $|xy| = |x|\,|y|$ and $|x| \ge 0$ (Basic properties of the absolute value).

[L5]

Finite sums are monotone in their terms and scale by a constant factor; the empty sum is $0$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L6]

For a series of nonnegative terms, convergence is equivalent to the range of the partial sums being bounded above, and then every partial sum is at most the sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

Direct comparison (If $0 \le a_k \le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$ , and divergence of $\sum a_k$ gives divergence of $\sum b_k$ ).

[L8]

If $\sum |x_k|$ converges then $\sum x_k$ converges (If $\sum |a_k|$ converges then $\sum a_k$ converges).

Proof

technique · direct

1.1

Both $L_a$ and $L_b$ are nonnegative, and $P_N \le L_a$ and $Q_m \le L_b$ for all $N$ and $m$ , the terms $|a_k|$ and $|b_j|$ being nonnegative.

givenL4L6

1.2

Put $\gamma_n := \sum_{k=0}^{n} |a_k|\,|b_{n-k}|$ , the Cauchy product of the sequences $(|a_k|)$ and $(|b_k|)$ ; every $\gamma_n$ is nonnegative.

givenL4L5

2.1

For every $n$ , $|c_n| = \bigl|\sum_{k=0}^{n} a_k b_{n-k}\bigr| \le \sum_{k=0}^{n} |a_k b_{n-k}| = \gamma_n$ .

step 1.2L3L4

2.2

Applying [L1] to $(|a_k|)$ and $(|b_k|)$ gives $\sum_{n<N} \gamma_n = \sum_{i<N} |a_i|\, Q_{N-i}$ for every $N$ .

step 1.2L1

3.1

Since $0 \le Q_{N-i} \le L_b$ and $|a_i| \ge 0$ , monotonicity and scaling give $\sum_{i<N} |a_i|\,Q_{N-i} \le \sum_{i<N} |a_i|\,L_b = L_b\,P_N \le L_b L_a$ for every $N$ .

step 1.1step 2.2L5

4.1

So $\sum \gamma_n$ is a series of nonnegative terms whose partial sums are bounded above by $L_a L_b$ ; it therefore converges, with sum at most $L_a L_b$ .

step 1.2step 3.1L6

5.1

By step 2.1 and comparison, $\sum |c_n|$ converges, and its sum is at most that of $\sum \gamma_n$ , hence at most $L_a L_b$ ; that is, $\sum c_n$ converges absolutely and satisfies the displayed bound.

step 2.1step 4.1L6L7

6.1

The hypotheses of Mertens' theorem hold, $\sum a_k$ converging absolutely and $\sum b_k$ converging by step 1.1 and [L8]; so $\sum c_n$ converges with sum $AB$ .

givenL2L8∎

Remarks

Only claim 1 of Mertens' theorem: if $\sum a_k$ converges absolutely to $A$ and $\sum b_k$ converges to $B$ , their Cauchy product converges to $AB$ is reused, and it is reused for a different pair of sequences. The identity there is proved for arbitrary real sequences and carries no convergence hypothesis, which is exactly what allows it to be applied here to $(|a_k|)$ and $(|b_k|)$ . Nothing about the absolute values is reproved.
The bound is not an equality. The sum of $\sum|c_n|$ can be strictly less than $L_a L_b$ , since cancellation inside each $c_n$ is invisible to $\gamma_n$ ; the inequality is all that is claimed and all that is needed.
Absolute convergence of both factors is a genuine strengthening. Mertens' theorem already gives $\sum c_n = AB$ with only one factor absolutely convergent; what the second hypothesis buys is that the product series is itself absolutely convergent, hence unconditionally convergent (Dirichlet's rearrangement theorem: an absolutely convergent series converges unconditionally, and every rearrangement of it has the same sum), so that its terms may be reordered in turn.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 71 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Cauchy product (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis, Chapter 4 (standard reference, not scraped)

If ∑ak\sum a_k∑ak​ and ∑bk\sum b_k∑bk​ both converge absolutely then their Cauchy product converges absolutely, with sum ABABAB