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$x_{k} = \sqrt{k}$ has $x_{k + 1} - x_{k} \to 0$ and is not Cauchy

Statement refuted

Refuted claim: a sequence of reals whose consecutive differences tend to $0$ is Cauchy (FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy, Limits and Cauchy sequences of reals).

The witness is $x_{k} = \sqrt{k}$ for $k \in N$ . Its consecutive differences satisfy

$x_{k + 1} - x_{k} = \sqrt{k + 1} - \sqrt{k} = \frac{1}{\sqrt{k + 1} + \sqrt{k}} ⟶ 0,$

while the sequence itself is unbounded, hence not Cauchy (Every Cauchy sequence of reals is bounded). The refutation is carried out in full in FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy; this item records the witness and adds the sharper statement that $\sqrt{k}$ diverges to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ).

Facts & Assumptions

Given: The sequence $(x_{k})$ of reals with $x_{k} : = \sqrt{k}$ , where $k$ denotes the canonical natural $k \cdot 1_{R}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L1]

The witness and its two properties: $x_{k + 1} - x_{k} = 1 / (\sqrt{k + 1} + \sqrt{k})$ tends to $0$ , and $(x_{k})$ is unbounded and not Cauchy (FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy).

[L2]

Square roots, and the factorisation $b^{2} - a^{2} = (b - a) (b + a)$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate, Integer powers $a^{m}$ ).

[L3]

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

[L4]

Canonical naturals: positive for $n \geq 1$ , and strictly increasing in the index (Canonical naturals are positive and strictly increasing); reciprocals of positives are positive and reciprocation reverses the order (Inverses of positives are positive, and reciprocation reverses order).

[L5]

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

[L6]

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

[L7]

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

[L8]

Divergence to $+ \infty$ : for every real $M$ there is $K$ with $x_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[L9]

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

Counterexample

technique · direct

1.1givenL1L7

The sequence $(x_{k})$ satisfies the hypothesis of the refuted claim, its consecutive differences tending to $0$ , and it is not Cauchy.

1.2givenL5L6

The failure is as strong as possible: $(x_{k})$ diverges to $+ \infty$ . Let $M \in R$ and put $M^{'} : = ∣ M ∣ \geq M$ , so $M^{'} \geq 0$ . By [L5] fix a natural $n \geq 1$ with $(M^{'})^{2} < n$ .

2.1step 1.1L1

It therefore refutes the claim: having null consecutive differences does not make a sequence Cauchy.

3.1step 1.2L2L3L4L8L9∎

For every $k \geq n$ : $(x_{k})^{2} = k \geq n > (M^{'})^{2}$ with $x_{k} \geq 0$ and $M^{'} \geq 0$ , so $x_{k} > M^{'} \geq M$ . Since $M$ was arbitrary, $x_{k} \to + \infty$ .

Remarks

The gaps are null and their sums are not. The differences are about $1 / (2 \sqrt{k})$ , so they tend to $0$ ; but they telescope, and $x_{m} - x_{k} = \sqrt{m} - \sqrt{k}$ is as large as one likes for $m$ large. The Cauchy condition constrains $∣ x_{m} - x_{k} ∣$ for all large pairs, and no hypothesis about consecutive pairs alone can deliver that.
What the correct hypothesis looks like. Geometric decay of the gaps, with a single ratio $c < 1$ at every index, does suffice (Contractive sequence: $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ for a fixed $0 < c < 1$ , Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ ), because then the telescoped sums are dominated by a convergent geometric bound. Merely shrinking gaps are not enough either, which is $x_{k + 1} = x_{k} + 1 / x_{k}$ from $x_{1} = 1$ has strictly decreasing consecutive gaps and diverges, so no uniform $c < 1$ exists.
The same sequence separates two notions that are easy to confuse. Its gaps are null, so it is "eventually almost constant" in a naive reading; it is nevertheless unbounded and divergent to $+ \infty$ . Nothing about the local behaviour of a sequence controls its global behaviour.

$x_{k} = \sqrt{k}$ has $x_{k + 1} - x_{k} \to 0$ and is not Cauchy

Statement refuted

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Counterexample

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Dependency tree · 2 levels

Sources

xk=kxk​=k​ has xk+1−xk→0xk+1​−xk​→0 and is not Cauchy

Statement refuted

Facts & Assumptions

Counterexample

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Dependency tree · 2 levels

Sources

$x_{k} = \sqrt{k}$ has $x_{k + 1} - x_{k} \to 0$ and is not Cauchy