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

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Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$

Statement

Let $(a_{k})$ and $(b_{k})$ be sequences of reals, let $A_{n} = \sum_{i < n} a_{i}$ and $B_{m} = \sum_{j < m} b_{j}$ be their partial sums (Series, partial sums, convergence and the sum, divergence, and the tail series), and let $(c_{n})$ be their Cauchy product, $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ). Then:

A finite identity, holding for arbitrary sequences. For every $N \in N$ , $\sum_{n < N} c_{n} = \sum_{i < N} a_{i} B_{N - i} .$
Mertens' theorem. If $\sum a_{k}$ converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index) and $\sum b_{k}$ converges, then $\sum a_{k}$ converges, say to $A$ , and, writing $B$ for the sum of $\sum b_{k}$ , the Cauchy product $\sum c_{n}$ converges with $\sum_{n = 0}^{\infty} c_{n} = A B .$

Claim 1 carries no hypothesis at all and is used again, for the sequences $(∣ a_{k} ∣)$ and $(∣ b_{k} ∣)$ , in If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ ; that is why it is stated as part of the theorem rather than buried in the proof.

The hypotheses are not symmetric, and that is the point. Only one of the two series is required to converge absolutely; the other need only converge. Requiring convergence of both and nothing more is not enough, as FALSE: the Cauchy product of two convergent series converges shows.

Facts & Assumptions

Given: Sequences $(a_{k})$ and $(b_{k})$ of reals, their partial sums $A_{n}$ and $B_{m}$ , and their 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]

Finite sums: $\sum_{k < 0} x_{k} = 0$ , $\sum_{k < n + 1} x_{k} = \sum_{k < n} x_{k} + x_{n}$ , and $\sum_{k = 0}^{n} x_{k} = \sum_{k < n + 1} x_{k}$ (Finite sums and finite products, by recursion).

[L2]

Finite sums are additive, are scaled by a constant factor, may be split at an intermediate index, and are monotone in their terms (Laws of finite sums and finite products).

[L3]

Partial sums of a series, and the meaning of its sum (Series, partial sums, convergence and the sum, divergence, and the tail series, Limits and Cauchy sequences of reals).

[L4]

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

[L5]

$∣ \sum_{k < n} x_{k} ∣ \leq \sum_{k < n} ∣ x_{k} ∣$ (Triangle inequality for finite sums).

[L6]

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

[L7]

For a series of nonnegative terms, every partial sum is at most the sum, and the partial sums converge to it (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L8]

A convergent sequence of reals is bounded (Every convergent sequence is bounded).

[L9]

Algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients, Limits and Cauchy sequences of reals).

[L10]

If $\sum ∣ x_{k} ∣$ converges then $\sum x_{k}$ converges; absolute convergence of $\sum a_{k}$ means convergence of $\sum ∣ a_{k} ∣$ (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges, Absolutely convergent and conditionally convergent series, and the general starting index).

Proof

technique · direct

1.1L1L2L4

Claim 1 holds, by induction on $N$ . At $N = 0$ both sides are empty sums, hence $0$ . Assume it at $N$ . By [L1], $\sum_{n < N + 1} c_{n} = \sum_{n < N} c_{n} + c_{N}$ and $c_{N} = \sum_{k < N + 1} a_{k} b_{N - k} = \sum_{k < N} a_{k} b_{N - k} + a_{N} b_{0}$ . On the other side, $\sum_{i < N + 1} a_{i} B_{N + 1 - i} = \sum_{i < N} a_{i} B_{N + 1 - i} + a_{N} B_{1}$ , where $B_{1} = b_{0}$ by [L1] and $B_{N + 1 - i} = B_{N - i} + b_{N - i}$ for every $i \leq N$ , again by [L1]; so additivity gives $\sum_{i < N + 1} a_{i} B_{N + 1 - i} = \sum_{i < N} a_{i} B_{N - i} + \sum_{i < N} a_{i} b_{N - i} + a_{N} b_{0}$ . Substituting the induction hypothesis into the first term and recognising the last two as $c_{N}$ closes the induction.

1.2givenL3L7L10

Assume the hypotheses of claim 2. Since $\sum ∣ a_{k} ∣$ converges, $\sum a_{k}$ converges; write $A$ for its sum, so $A_{n} \to A$ , and write $L$ for the sum of $\sum ∣ a_{k} ∣$ , so that $P_{N} : = \sum_{k < N} ∣ a_{k} ∣$ satisfies $P_{N} \leq L$ for every $N$ and $P_{N} \to L$ .

1.3givenL3L8L9choose

Write $B$ for the sum of $\sum b_{k}$ and $β_{m} : = B_{m} - B$ , so that $β_{m} \to 0$ ; being convergent, $(β_{m})$ is bounded, and we fix a real $C \geq 1$ with $∣ β_{m} ∣ \leq C$ for every $m$ .

2.1step 1.1step 1.3L2

By claim 1 and additivity, for every $N$ , $\sum_{n < N} c_{n} = \sum_{i < N} a_{i} (B + β_{N - i}) = B A_{N} + R_{N}$ , where $R_{N} : = \sum_{i < N} a_{i} β_{N - i}$ .

2.2step 1.2step 1.3choose

Let $ε > 0$ be real. Since $P_{N} \to L$ , fix $M \in N$ with $L - P_{M} < ε / (2 C)$ ; since $β_{m} \to 0$ , fix $K \in N$ with $∣ β_{m} ∣ < ε / (2 (L + 1))$ for all $m \geq K$ . Both quotients are legitimate, $C \geq 1$ and $L + 1 \geq 1$ being positive.

3.1step 2.1L2L5L6

For $N \geq M + K$ , the triangle inequality and splitting at $M$ give $∣ R_{N} ∣ \leq \sum_{i < N} ∣ a_{i} ∣ ∣ β_{N - i} ∣ = \sum_{i < M} ∣ a_{i} ∣ ∣ β_{N - i} ∣ + \sum_{i = M}^{N - 1} ∣ a_{i} ∣ ∣ β_{N - i} ∣$ .

4.1step 2.2step 3.1step 1.2L2

In the first of those sums $i < M$ and $N \geq M + K$ , so $N - i > N - M \geq K$ and in particular $N - i \geq K$ , whence $∣ β_{N - i} ∣ < ε / (2 (L + 1))$ ; monotonicity of finite sums then bounds it by $ε P_{M} / (2 (L + 1)) \leq ε L / (2 (L + 1)) < ε / 2$ .

4.2step 2.2step 3.1step 1.2step 1.3L2

In the second sum every factor $∣ β_{N - i} ∣$ is at most $C$ , so it is bounded by $C \sum_{i = M}^{N - 1} ∣ a_{i} ∣ = C (P_{N} - P_{M}) \leq C (L - P_{M}) < ε / 2$ .

5.1step 3.1step 4.1step 4.2L3

Hence $∣ R_{N} ∣ < ε$ for every $N \geq M + K$ ; as $ε > 0$ was arbitrary, $R_{N} \to 0$ .

6.1step 1.2step 2.1step 5.1L9∎

By step 2.1, step 1.2 and step 5.1 the partial sums of $\sum c_{n}$ satisfy $\sum_{n < N} c_{n} = B A_{N} + R_{N} \to B A + 0 = A B$ , so $\sum c_{n}$ converges with sum $A B$ , which is claim 2.

Remarks

Where absolute convergence of $\sum a_{k}$ is used. Twice, and both times to control a tail of $\sum ∣ a_{k} ∣$ : in step 2.2, to make the far block of the splitting small uniformly in $N$ , and in step 4.1, where $P_{M} \leq L$ bounds the near block. Mere convergence of $\sum a_{k}$ gives no such control, since the tail of a conditionally convergent series is small only after cancellation, and the factors $β_{N - i}$ destroy the cancellation.
The identity of claim 1 is a rectangle folded into a triangle. It says that summing the products $a_{i} b_{j}$ over the triangle $i + j < N$ by antidiagonals gives the same result as summing them row by row, $\sum_{i < N} a_{i} \sum_{j < N - i} b_{j}$ . The induction proves exactly that, and it needs no hypothesis because both sides are finite sums.
Abel's stronger theorem is not available here. If $\sum a_{k}$ , $\sum b_{k}$ and $\sum c_{n}$ all converge, then the sum of $\sum c_{n}$ is $A B$ without any absolute convergence; but the standard proof runs through power series and Abel's limit theorem, which are later in the reading order. Mertens' theorem is what this page can prove, and its hypotheses are what If $\sum a_{k}$ and $\sum b_{k}$ both converge absolutely then their Cauchy product converges absolutely, with sum $A B$ inherits.

Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Mertens' theorem: if ∑ak∑ak​ converges absolutely to AA and ∑bk∑bk​ converges to BB, their Cauchy product converges to ABAB

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$