False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: the Cauchy product of two convergent series converges

Statement

False claim: if $\sum a_{k}$ and $\sum b_{k}$ both converge (Series, partial sums, convergence and the sum, divergence, and the tail series) then their Cauchy product $\sum c_{n}$ converges (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ).

What is true is Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ , which requires one of the two factors to converge absolutely. Convergence of both is not enough, and the standard witness is a single series multiplied by itself.

Let $(ε_{k})$ be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ) and put

$a_{k} = b_{k} : = \frac{ε_{k}}{\sqrt{ι (k + 1)}} (k \in N),$

with $\sqrt{}$ the nonnegative square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ) and $ι (k + 1)$ the canonical natural, positive for every $k$ (Canonical naturals are positive and strictly increasing). Then $\sum a_{k}$ converges, by the alternating series test, while the Cauchy product satisfies

$∣ c_{n} ∣ \geq \frac{2 ι (n + 1)}{ι (n + 2)} \geq 1 for every n \in N,$

so $(c_{n})$ does not converge to $0$ and $\sum c_{n}$ diverges (If a series converges then its terms tend to $0$ ).

Facts & Assumptions

Given: The alternating sequence $(ε_{k})$ , the sequence $β_{k} : = 1 / \sqrt{ι (k + 1)}$ , the sequence $a_{k} = b_{k} = ε_{k} β_{k}$ , and their Cauchy product $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ (The Cauchy product of two series: $c_{n} = \sum_{k = 0}^{n} a_{k} b_{n - k}$ ).

[A1]

The refuted claim: the Cauchy product of two convergent series of reals converges.

[L1]

The alternating sequence: $ε_{0} = 1$ , $ε_{k + 1} = - ε_{k}$ , and $∣ ε_{k} ∣ = 1$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

Square roots: every $t \geq 0$ has a unique $\sqrt{t} \geq 0$ with $(\sqrt{t})^{2} = t$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L3]

The canonical naturals: $ι (n) > 0$ for $n \geq 1$ , $ι$ is strictly increasing, and $ι (m + n) = ι (m) + ι (n)$ (Canonical naturals are positive and strictly increasing).

[L4]

If $0 < u < v$ then $0 < 1 / v < 1 / u$ (Inverses of positives are positive, and reciprocation reverses order).

[L5]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / ι (n) < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

The alternating series test (The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ , Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L7]

AM-GM for two nonnegative reals, in the product form: $u v \leq ((u + v) / 2)^{2}$ (The arithmetic mean, geometric mean inequality).

[L8]

Finite sums: the sum of a constant, monotonicity in the terms, and $\sum_{k = 0}^{n} x_{k} = \sum_{k < n + 1} x_{k}$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion).

[L9]

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

[L10]

If $\sum x_{n}$ converges then $x_{n} \to 0$ (If a series converges then its terms tend to $0$ , Limits and Cauchy sequences of reals).

[L11]

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

[L12]

Mertens' theorem, which requires one factor to converge absolutely (Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ ).

Refutation

technique · direct

1.1

Square roots are strictly increasing on the nonnegative reals: if $0 \leq u < v$ and $\sqrt{u} \geq \sqrt{v}$ then $u = (\sqrt{u})^{2} \geq (\sqrt{v})^{2} = v$ , which is false; so $\sqrt{u} < \sqrt{v}$ . Also $\sqrt{u v} = \sqrt{u} \sqrt{v}$ for $u, v \geq 0$ , since $(\sqrt{u} \sqrt{v})^{2} = u v$ and $\sqrt{u} \sqrt{v} \geq 0$ , and $\sqrt{t^{2}} = t$ for $t \geq 0$ .

1.2

An induction on $j$ gives $ε_{m} ε_{j} = ε_{m + j}$ for all $m, j$ : at $j = 0$ this is $ε_{m} \cdot 1 = ε_{m}$ , and $ε_{m} ε_{j + 1} = ε_{m} (- ε_{j}) = - ε_{m + j} = ε_{m + j + 1}$ .

L1L11

2.1

Each $β_{k} = 1 / \sqrt{ι (k + 1)}$ is a positive real, and $(β_{k})$ is nonincreasing, since $0 < ι (k + 1) < ι (k + 2)$ gives $0 < \sqrt{ι (k + 1)} < \sqrt{ι (k + 2)}$ and inverting reverses the inequality.

step 1.1L3L4

2.2

$(β_{k})$ converges to $0$ : given a rational $ε > 0$ , fix a natural $n \geq 1$ with $1 / ι (n) < ε^{2}$ ; then for $k \geq n$ one has $ι (k + 1) \geq ι (n) > 1 / ε^{2} = (1 / ε)^{2}$ , so $\sqrt{ι (k + 1)} > 1 / ε$ and $β_{k} < ε$ .

step 1.1L3L4L5

2.3

For $k \leq n$ , [L7] applied to $u = ι (k + 1)$ and $v = ι (n - k + 1)$ , whose sum is $ι (n + 2)$ by [L3], gives $ι (k + 1) ι (n - k + 1) \leq (ι (n + 2) / 2)^{2}$ ; taking square roots and using step 1.1, $\sqrt{ι (k + 1)} \sqrt{ι (n - k + 1)} \leq ι (n + 2) / 2$ .

step 1.1L3L7

3.1

By the alternating series test $\sum a_{k} = \sum ε_{k} β_{k}$ converges; the same series is taken as both factors.

step 2.1step 2.2L6

3.2

Hence for every $n$ and every $k \leq n$ , $a_{k} b_{n - k} = ε_{k} ε_{n - k} β_{k} β_{n - k} = ε_{n} β_{k} β_{n - k}$ , so $c_{n} = ε_{n} \sum_{k = 0}^{n} β_{k} β_{n - k}$ and $∣ c_{n} ∣ = \sum_{k = 0}^{n} β_{k} β_{n - k}$ , the terms being positive.

step 2.1step 1.2L1L8L9

3.3

Inverting, $β_{k} β_{n - k} \geq 2 / ι (n + 2)$ for every $k \leq n$ .

step 2.3L4

4.1

Summing the $n + 1$ terms and using monotonicity of finite sums and the sum of a constant, $∣ c_{n} ∣ \geq ι (n + 1) \cdot 2 / ι (n + 2) = 2 ι (n + 1) / ι (n + 2)$ .

step 3.2step 3.3L8

5.1

Moreover $2 ι (n + 1) = ι (2 n + 2) \geq ι (n + 2)$ , since $2 n + 2 \geq n + 2$ and $ι$ is increasing; so $∣ c_{n} ∣ \geq 1$ for every $n$ .

step 4.1L3

6.1

The sequence $(c_{n})$ therefore does not converge to $0$ : the tolerance $ε = 1$ admits no index $K$ with $∣ c_{n} - 0 ∣ < 1$ for all $n \geq K$ . Hence $\sum c_{n}$ diverges.

step 5.1L10

7.1

So both factors converge while their Cauchy product diverges, and the claim [A1] is false; what is true is [L12], which asks one factor to converge absolutely, and this witness cannot satisfy that hypothesis, since otherwise Mertens' theorem would make $\sum c_{n}$ convergent, contrary to step 6.1.

step 3.1step 6.1A1L12∎

Remarks

The lower bound is not merely nonzero: it grows to $2$ . Step 4.1 gives $∣ c_{n} ∣ \geq 2 ι (n + 1) / ι (n + 2) = 2 - 2 / ι (n + 2)$ , and that bound increases to $2$ ; so the terms of the Cauchy product do not shrink at all, and the divergence is detected by the crudest test available. What the size of $∣ c_{n} ∣$ itself tends to is not determined here and is not needed.
Where the failure comes from. In $c_{n}$ every one of the $n + 1$ products $a_{k} b_{n - k}$ carries the same sign $ε_{n}$ , so no cancellation occurs within $c_{n}$ : the alternation that makes each factor converge is exactly what aligns the terms of the product. Absolute convergence of one factor, as in Mertens' theorem: if $\sum a_{k}$ converges absolutely to $A$ and $\sum b_{k}$ converges to $B$ , their Cauchy product converges to $A B$ , prevents this by making the total mass finite.
The claim becomes true under other hypotheses. If all three series $\sum a_{k}$ , $\sum b_{k}$ and $\sum c_{n}$ are assumed to converge, then the sum of the product is the product of the sums; but that theorem is proved through power series and Abel's limit theorem, which are later in the reading order. The companion examples page records the same witness from the other side.

Depends on

Used by

The Cauchy product of ∑_k ≥ 0 (-1)ᵏ/√k+1 with itself has |cₙ| ≥ 1 for every n, so it diverges Counterexample

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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)
Colorado State University, MATH 171 Homework 4 Solutions (standard reference, not scraped)