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

1.1

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

givenL2

2.1

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

step 1.1L2L6

2.2

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: $(g_{j + 1})^{2} = (g_{j} + 1 / g_{j})^{2} = (g_{j})^{2} + 2 + 1 / (g_{j})^{2} > (g_{j})^{2} + 2 \geq (1 + 2 j) + 2 = 1 + 2 (j + 1)$ .

step 1.1L1L2L3L4

3.1

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.

step 1.1step 2.1L2L5

3.2

$(g_{j})$ 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 $g_{k} \geq g_{n} > M$ .

step 2.1step 2.2L3L4L5L7

4.1

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

step 3.2L7

5.1

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.

step 4.1L8

6.1

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.

step 3.1step 3.2step 5.1L8∎

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.

Depends on

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Sources

Contraction mapping (Wikipedia) (standard reference, not scraped)
Cauchy sequence (Wikipedia) (standard reference, not scraped)
R. Bartle and D. Sherbert, Introduction to Real Analysis, 4th ed., §3.5 (standard reference, not scraped)

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