CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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$x_{k + 1} = x_{k} + 1 / x_{k}$ from $x_{1} = 1$ has strictly decreasing consecutive gaps and diverges, so no uniform $c < 1$ exists

Statement refuted

Refuted claim: a sequence whose consecutive gaps are strictly decreasing,

$∣ x_{k + 2} - x_{k + 1} ∣ < ∣ x_{k + 1} - x_{k} ∣ for every k,$

is contractive, or at least converges (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 $k \geq 1$ ).

The witness is $x_{1} = 1$ , $x_{k + 1} = x_{k} + 1 / x_{k}$ . Its gaps are $x_{k + 1} - x_{k} = 1 / x_{k}$ , strictly decreasing because $(x_{k})$ is strictly increasing; and the sequence diverges to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ). Since a contractive sequence converges (Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ ), no contraction constant $c < 1$ can exist for it: the ratios of consecutive gaps are all below $1$ but have no bound below $1$ that works at every index.

Indexing. Written on $N$ the sequence is $(g_{j})$ with $g_{0} = 1$ and $g_{j + 1} = g_{j} + 1 / g_{j}$ , and $x_{k} : = g_{k - 1}$ for $k \geq 1$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Facts & Assumptions

Given: The set $A : = {u \in R : u \geq 1}$ , the element $1 \in A$ , and the function $f : A \to A$ with $f (u) = u + 1 / u$ , which lands in $A$ because $u \geq 1 > 0$ gives $1 / u > 0$ and hence $f (u) > u \geq 1$ ; by the recursion theorem (The recursion theorem) the unique $g : N \to A$ with $g_{0} = 1$ and $g_{j + 1} = f (g_{j})$ ; and the gaps $d_{j} : = g_{j + 1} - g_{j}$ .

[L1]

Recursion theorem (The recursion theorem) and induction principle (The principle of mathematical induction).

[L2]

Order and arithmetic: $1 > 0$ , so $2 = 1 + 1 > 0$ ; sums of positives are positive; adding a constant preserves the order; a positive has a positive inverse, and $0 < a < b$ gives $0 < 1 / b < 1 / a$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Inverses of positives are positive, and reciprocation reverses order, Ordered field, Complete ordered field (least-upper-bound property)).

[L3]

Powers: $(u + v)^{2} = u^{2} + 2 u v + v^{2}$ and $u \cdot (1 / u) = 1$ , so $(u + 1 / u)^{2} = u^{2} + 2 + 1 / u^{2}$ ; and for $a, b \geq 0$ , $a < b$ exactly when $a^{2} < b^{2}$ (Integer powers $a^{m}$ , 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); the Archimedean property (Every complete ordered field is Archimedean).

[L5]

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

[L6]

Monotone sequences, with consecutive comparisons sufficing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L7]

Divergence to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ); a convergent sequence is bounded (Every convergent sequence is bounded); convergence (Limits and Cauchy sequences of reals).

[L8]

Contractive sequences and their convergence: a contractive sequence, with a constant $c$ satisfying $0 < c < 1$ at every index, converges (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 $k \geq 1$ ).

Counterexample

technique · direct

1.1givenL2

Every term satisfies $g_{j} \geq 1 > 0$ , since $g$ takes values in $A$ by construction.

2.1step 1.1L2L6

The sequence is strictly increasing: $d_{j} = g_{j + 1} - g_{j} = 1 / g_{j} > 0$ , and consecutive comparisons give strict increase.

2.2step 1.1L1L2L3L4

By induction, $(g_{j})^{2} \geq 1 + 2 j$ for every $j$ , where $j$ denotes the canonical natural. Base: $(g_{0})^{2} = 1 = 1 + 2 \cdot 0$ . Step: $(g_{j + 1})^{2} = (g_{j} + 1 / g_{j})^{2} = (g_{j})^{2} + 2 + 1 / (g_{j})^{2} > (g_{j})^{2} + 2 \geq (1 + 2 j) + 2 = 1 + 2 (j + 1)$ .

3.1step 1.1step 2.1L2L5

The gaps are strictly decreasing: $0 < g_{j} < g_{j + 1}$ gives $0 < 1 / g_{j + 1} < 1 / g_{j}$ , that is $0 < d_{j + 1} < d_{j}$ ; and $∣ d_{j + 1} ∣ < ∣ d_{j} ∣$ since both are positive.

3.2step 2.1step 2.2L3L4L5L7

$(g_{j})$ diverges to $+ \infty$ . Let $M \in R$ and put $M^{'} : = ∣ M ∣ \geq M$ , so $M^{'} \geq 0$ . By [L4] fix a natural $n \geq 1$ with $(M^{'})^{2} < n$ . Then $(g_{n})^{2} \geq 1 + 2 n > n > (M^{'})^{2}$ , and since $g_{n} \geq 0$ and $M^{'} \geq 0$ this gives $g_{n} > M^{'} \geq M$ ; for every $k \geq n$ strict increase gives $g_{k} \geq g_{n} > M$ .

4.1step 3.2L7

$(g_{j})$ does not converge: a convergent sequence is bounded, whereas step 3.2 exhibits terms above every real.

5.1step 4.1L8

No $c$ with $0 < c < 1$ is a contraction constant for $(g_{j})$ : if one were, the sequence would be contractive and would converge, contradicting step 4.1.

6.1step 3.1step 3.2step 5.1L8∎

So $(g_{j})$ , that is $(x_{k})$ , has strictly decreasing consecutive gaps by step 3.1, diverges to $+ \infty$ by step 3.2, and admits no uniform $c < 1$ by step 5.1: strictly decreasing gaps neither make a sequence contractive nor make it converge.

Remarks

The gaps are not merely decreasing, they are null. Since $d_{j} = 1 / g_{j}$ and $g_{j} \to + \infty$ , for any real $ε > 0$ the terms eventually exceed $1 / ε$ , so $0 < d_{j} < ε$ eventually (Inverses of positives are positive, and reciprocation reverses order). So this sequence is also a witness for FALSE: if $∣ x_{k + 1} - x_{k} ∣ \to 0$ then $(x_{k})$ is Cauchy, alongside $\sqrt{k}$ ; the two are close relatives, since $(g_{j})^{2} \geq 1 + 2 j$ says $g_{j}$ grows at least like $\sqrt{2 j}$ .
What the uniform constant is really asking for. The ratio of consecutive gaps here is $d_{j + 1} / d_{j} = g_{j} / g_{j + 1}$ , which is below $1$ at every index and approaches $1$ as $j$ grows. A contraction constant would have to sit strictly between all of those ratios and $1$ , and there is no room: the supremum of the ratios is $1$ itself. This is the precise sense in which Contractive sequence: $∣ x_{k + 2} - x_{k + 1} ∣ \leq c ∣ x_{k + 1} - x_{k} ∣$ for a fixed $0 < c < 1$ asks for more than "each gap smaller than the last".
The comparison with a genuine contraction. In The sequence $x_{k + 1} = (x_{k} + 1) / 3$ is contractive with $c = 1 / 3$ and converges to $1 / 2$ the ratio is exactly $1 / 3$ at every index, so the constant exists and is optimal, and the error bound of Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ applies. The difference between the two examples is not the speed at which the gaps shrink at any given index but whether the shrinking is uniform.

$x_{k + 1} = x_{k} + 1 / x_{k}$ from $x_{1} = 1$ has strictly decreasing consecutive gaps and diverges, so no uniform $c < 1$ exists

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

xk+1=xk+1/xkxk+1​=xk​+1/xk​ from x1=1x1​=1 has strictly decreasing consecutive gaps and diverges, so no uniform c<1c<1 exists

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

$x_{k + 1} = x_{k} + 1 / x_{k}$ from $x_{1} = 1$ has strictly decreasing consecutive gaps and diverges, so no uniform $c < 1$ exists