Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every contractive sequence is Cauchy, hence converges, with error bound ∣x−xk∣≤ck−1∣x2−x1∣/(1−c) for k≥1

Statement

Let (xk) be a contractive sequence of reals with contraction constant c, so 0<c<1 and ∣xk+2−xk+1∣≤c ∣xk+1−xk∣ for every k∈N (Contractive sequence: ∣xk+2−xk+1∣≤c ∣xk+1−xk∣ for a fixed 0<c<1). Then:

  1. Geometric decay of the gaps. For every k≥1, ∣xk+1−xk∣  ≤  c k−1 ∣x2−x1∣.
  2. Convergence. (xk) is Cauchy (Limits and Cauchy sequences of reals) and therefore converges to some x∈R (The Cauchy criterion from the least-upper-bound property: in a complete ordered field every Cauchy sequence converges).
  3. Error bound. For every k≥1, ∣x−xk∣  ≤  c k−1 ∣x2−x1∣1−c.

The restriction k≥1 in claim 3 is a hypothesis, not a convention. The displayed bound is false at k=0, even though c−1 is defined (Integer powers am). Take c=1/2 and the sequence x0=0, xk=1 for all k≥1: it is contractive with that c, its limit is x=1, the right-hand side at k=0 is c−1 ∣x2−x1∣/(1−c)=2⋅0/(1/2)=0, and the left-hand side is ∣x−x0∣=1. The classical statement of this theorem is written for sequences indexed from 1, where the question does not arise; this library indexes from 0 (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so the hypothesis is stated.

Facts & Assumptions

Given: A sequence (xk) of reals and a real c with 0<c<1 such that ∣xk+2−xk+1∣≤c ∣xk+1−xk∣ for every k∈N; the abbreviations D:=∣x2−x1∣≥0 and A:=D/(1−c), which is defined and ≥0 since 1−c>0.

[L2]
[L3]

Integer powers: a0=1, ak+1=aka; and the law am+n=aman (Integer powers am, Laws of integer exponents).

[L4]

Powers and order: a≥0 gives an≥0; 1n=1 for every n (Monotonicity of x↦xn and of n↦an).

[L5]

Absolute value: ∣t∣≥0, ∣t∣=∣−t∣, and ∣t∣≤u exactly when −u≤t≤u (Basic properties of the absolute value).

[L6]

Multiplying inequalities of nonnegatives: 0≤a≤b and 0≤e≤f give ae≤bf (Multiplying inequalities of positives).

[L8]

Finite sums, their notation ∑i<n, and their laws: additivity, scaling, monotonicity, and telescoping ∑i<n(ti+1−ti)=tn−t0 for any sequence (ti) (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L9]

Triangle inequality for finite sums: ∣∑i<nai∣≤∑i<n∣ai∣ (Triangle inequality for finite sums).

[L10]

Factorisation: 1−c n=(1−c)∑i<nc i, the case b=1, a=c of bn−an=(b−a)∑i=0n−1aib n−1−i together with 1m=1; at n=0 both sides are 0 (Factorisation of bn−an, and the resulting Lipschitz estimate, Monotonicity of x↦xn and of n↦an).

[L12]

Cauchy condition and convergence; it suffices to test a real ε>0, since every positive rational is a positive real (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L14]

Limits: a sequence and each of its tails converge to the same limit (Convergence depends only on the tail); the algebra of limits (Algebra of limits: sums, scalar multiples, products and quotients); compatibility of the absolute value with limits (The absolute value is compatible with limits); and preservation of non-strict inequalities in the limit (Limits preserve non-strict inequalities).

[L15]

The order on N is total, so any two indices are comparable (≤ is a linear order on N).

Proof

technique · induction
1.1

Base case of claim 1, at k=1: ∣x2−x1∣≤c 0∣x2−x1∣=D, since c 0=1.

givenL3base
1.2

Inductive hypothesis: fix j∈N and assume ∣xj+2−xj+1∣≤c jD.

ih
1.3

By [L10], (1−c)∑i<nc i=1−c n≤1, since c n≥0; dividing by 1−c>0 gives ∑i<nc i≤1/(1−c).

givenL4L6L7L10
1.4

Let ε>0 be an arbitrary real and put δ:=ε/(2(A+1))>0, which is defined since A+1>0. By [L11] fix J∈N with c j<δ for every j≥J.

givenL4L7L11L12choose
2.1

Successor step: contractivity at the index j+1 gives ∣xj+3−xj+2∣≤c ∣xj+2−xj+1∣≤c⋅c jD=c j+1D, the middle inequality by the inductive hypothesis multiplied by c>0.

step 1.2L1L3L6
3.1

By the induction principle, ∣xj+2−xj+1∣≤c jD for every j∈N; writing k=j+1 this is claim 1: ∣xk+1−xk∣≤c k−1D for every k≥1.

step 1.1step 2.1L2
4.1

Fix k≥1 and n∈N, and put ti:=xk+i. Telescoping gives xk+n−xk=tn−t0=∑i<n(ti+1−ti), so ∣xk+n−xk∣≤∑i<n∣xk+i+1−xk+i∣.

step 3.1L8L9
4.2

Each summand obeys claim 1 at the index k+i≥1: ∣xk+i+1−xk+i∣≤c k+i−1D=c k−1c iD.

step 3.1L3
5.1

Summing the bound of step 4.2 over i<n, by monotonicity and scaling of finite sums, ∣xk+n−xk∣≤c k−1D∑i<nc i.

step 4.1step 4.2L8
6.1

Combining steps 5.1 and 1.3: for every k≥1 and every n∈N, ∣xk+n−xk∣≤c k−1D/(1−c)=c k−1A.

step 5.1step 1.3L4L6
7.1

For all indices l,k≥J+1: by comparability one of them is the smaller, say k, and writing l=k+n step 6.1 gives ∣xl−xk∣≤c k−1A≤c k−1(A+1)≤δ(A+1)=ε/2<ε, using k−1≥J; the case l<k follows since ∣xl−xk∣=∣xk−xl∣.

step 6.1step 1.4L5L6L15
8.1

The real ε>0 was arbitrary and the index J+1 was produced from it, so (xk) is Cauchy, and therefore converges to some x∈R: this is claim 2.

step 7.1L12L13
9.1

Fix k≥1. The k-th tail n↦xk+n converges to x, so as n ranges over N the sequence n↦xk+n−xk converges to x−xk, so n↦∣xk+n−xk∣ converges to ∣x−xk∣; the constant sequence with value c k−1A converges to c k−1A, and step 6.1 compares the two at every n.

step 6.1step 8.1L14
10.1

Preservation of non-strict inequalities in the limit therefore gives ∣x−xk∣≤c k−1A=c k−1∣x2−x1∣/(1−c) for every k≥1, which is claim 3; claims 1, 2 and 3 are thus all established.

step 3.1step 8.1step 9.1L14discharge-induction∎

Remarks

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