False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-06 (claude-sonnet-5)

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FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy

Statement

False claim: if $(x_{k})$ is a sequence of reals whose consecutive differences tend to $0$ , that is $∣ x_{k + 1} - x_{k} ∣ \to 0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals), then $(x_{k})$ is Cauchy.

The claim is the tempting misreading of the Cauchy condition. Being Cauchy requires $∣ x_{m} - x_{k} ∣$ to be small for all large $m$ and $k$ ; the hypothesis above controls only the case $m = k + 1$ , and finitely many small steps still accumulate without bound.

The witness is $x_{k} = \sqrt{k}$ , refuted below and recorded separately as the named counterexample of the companion page. Its consecutive differences are $\sqrt{k + 1} - \sqrt{k} = 1 / (\sqrt{k + 1} + \sqrt{k})$ , which tend to $0$ , while the sequence itself is unbounded and so cannot be Cauchy (Every Cauchy sequence of reals is bounded).

What is true in this direction is Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ : if the differences shrink geometrically, with a single ratio $c < 1$ working at every index (Contractive sequence: $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ for a fixed $0 < c < 1$ ), then the sequence is Cauchy. The gap between the two hypotheses is exactly the uniform ratio.

Facts & Assumptions

Given: The sequence $(x_{k})$ of reals with $x_{k} : = \sqrt{k}$ , where $k$ denotes the canonical natural $k \cdot 1_{R}$ and $\sqrt{\cdot}$ the nonnegative square root (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

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

[L2]

Powers and order: for $a, b \geq 0$ and $n \geq 1$ , $a < b$ exactly when $a^{n} < b^{n}$ , and $a \leq b$ exactly when $a^{n} \leq b^{n}$ ; and $a \geq 0$ gives $a^{n} \geq 0$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L3]

Factorisation at $n = 2$ : $b^{2} - a^{2} = (b - a) (b + a)$ (Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate); and $(u v)^{n} = u^{n} v^{n}$ , so $(1 / t)^{2} = 1 / t^{2}$ for $t \neq 0$ (Laws of integer exponents).

[L4]

Canonical naturals: $n \cdot 1_{R} > 0$ for $n \geq 1$ , $0 \cdot 1_{R} = 0$ , and $m \mapsto m \cdot 1_{R}$ is strictly increasing (Canonical naturals are positive and strictly increasing).

[L5]

Reciprocals: $a > 0$ gives $1 / a > 0$ , and $0 < a < b$ gives $0 < 1 / b < 1 / a$ (Inverses of positives are positive, and reciprocation reverses order).

[L6]

Archimedean property, in both forms: for every real $x$ there is a natural $n \geq 1$ with $x < n$ , and for every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (Every complete ordered field is Archimedean, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L7]

Absolute value: $∣ t ∣ \geq 0$ , $∣ t ∣ \geq t$ , and $∣ t ∣ = t$ for $t \geq 0$ (Basic properties of the absolute value).

[L8]

Every Cauchy sequence of reals is bounded (Every Cauchy sequence of reals is bounded).

[L9]

Convergence to $0$ , boundedness, and the Cauchy condition; it suffices to test a real $ε > 0$ (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L10]

Trichotomy of the order on $R$ (Complete ordered field (least-upper-bound property), Ordered field).

[L11]

The refuted claim: a sequence of reals whose consecutive differences tend to $0$ is Cauchy.

Refutation

technique · direct

1.1givenL1L2L4

Each $x_{k} = \sqrt{k}$ is defined and $\geq 0$ , since the canonical natural $k$ satisfies $k \geq 0$ ; and $x_{k + 1} > 0$ for every $k$ , since $k + 1 \geq 1$ gives $k + 1 > 0$ and hence $\sqrt{k + 1} > 0$ .

1.2givenL1L2L4L6L7

$(x_{k})$ is not bounded. Let $M \in R$ and put $M^{'} : = ∣ M ∣ \geq M$ , so $M^{'} \geq 0$ . By [L6] fix a natural $n \geq 1$ with $(M^{'})^{2} < n$ . Then $(x_{n})^{2} = n > (M^{'})^{2}$ with $x_{n} \geq 0$ and $M^{'} \geq 0$ , so $x_{n} > M^{'} \geq M$ , and $∣ x_{n} ∣ = x_{n} > M$ .

2.1step 1.1L1L3L5

For every $k$ : $(\sqrt{k + 1} - \sqrt{k}) (\sqrt{k + 1} + \sqrt{k}) = (\sqrt{k + 1})^{2} - (\sqrt{k})^{2} = (k + 1) - k = 1$ , and $\sqrt{k + 1} + \sqrt{k} \geq \sqrt{k + 1} > 0$ , so $\sqrt{k + 1} - \sqrt{k} = 1 / (\sqrt{k + 1} + \sqrt{k}) > 0$ .

2.2step 1.2L8L9L10

A Cauchy sequence of reals is bounded, so an unbounded sequence is not Cauchy; by step 1.2 no real $M$ bounds $(∣ x_{k} ∣)$ , so $(x_{k})$ is not Cauchy.

3.1step 2.1L5L7

Hence $∣ x_{k + 1} - x_{k} ∣ = \sqrt{k + 1} - \sqrt{k} = 1 / (\sqrt{k + 1} + \sqrt{k}) \leq 1 / \sqrt{k + 1}$ , the last inequality because $0 < \sqrt{k + 1} \leq \sqrt{k + 1} + \sqrt{k}$ .

4.1step 3.1L4L5L6choose

Let $ε > 0$ be real. By [L6] fix a natural $n \geq 1$ with $1 / n < ε^{2}$ . For every $k \geq n$ we have $k + 1 > n > 0$ , so $1 / (k + 1) < 1 / n < ε^{2}$ .

5.1step 3.1step 4.1L2L3L5

Taking square roots in step 4.1: $(1 / \sqrt{k + 1})^{2} = 1 / (k + 1) < ε^{2}$ with both $1 / \sqrt{k + 1} \geq 0$ and $ε > 0$ , so $1 / \sqrt{k + 1} < ε$ , and therefore $∣ x_{k + 1} - x_{k} ∣ < ε$ for every $k \geq n$ .

6.1step 5.1L9L11

The real $ε > 0$ was arbitrary, so the consecutive differences of $(x_{k})$ tend to $0$ : the sequence satisfies the hypothesis of the claim.

7.1step 6.1step 2.2L11∎

The sequence $x_{k} = \sqrt{k}$ therefore has consecutive differences tending to $0$ and is not Cauchy: the claim is false.

Remarks

The failure is not marginal. The witness does not merely fail to be Cauchy; it is unbounded, and indeed $\sqrt{k} \to + \infty$ . The consecutive differences are of size roughly $1 / (2 \sqrt{k})$ , so they are null, but their partial sums telescope to $\sqrt{m} - \sqrt{k}$ , which is large when $m$ is much larger than $k$ . Nothing about "small steps" constrains what many steps accumulate to.
The repair is a uniform ratio, not a faster rate. It is tempting to think that a fast enough decay of the gaps would suffice, and in a sense that is true, since summability of the gaps implies Cauchy; but the hypothesis available in practice is the contractive one, a single $c < 1$ with $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ , and that is what Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ consumes. Merely having each gap smaller than the last is not enough either, which is the separate witness $x_{k + 1} = x_{k} + 1 / x_{k}$ from $x_{1} = 1$ has strictly decreasing consecutive gaps and diverges, so no uniform $c < 1$ exists ↗.
Two of the three false statements on this page have the same shape. A condition that looks like the Cauchy condition, but at only one pair of indices per step, is not the Cauchy condition. The other one is FALSE: a sequence with a convergent subsequence is bounded (the converse of Bolzano-Weierstrass), where a condition holding along one subsequence is mistaken for a condition on the sequence.
The witness is recorded as the named counterexample $x_{k} = \sqrt{k}$ has $x_{k + 1} - x_{k} \to 0$ and is not Cauchy ↗, which adds the sharper statement that $\sqrt{k}$ diverges to $+ \infty$ .

FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: if ∣xk+1−xk∣→0∣xk+1​−xk​∣→0 then (xk)(xk​) is Cauchy

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy