ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

How statement and proof provenance work

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.

Changing the first hundred terms leaves convergence to $0$ and boundedness intact

Example

Let $u_{k} : = (k + 1)^{- 1}$ and let $(x_{k})$ be $u$ modified in its first hundred terms: $x_{k} : = k + 1$ for $k < 100$ and $x_{k} : = (k + 1)^{- 1}$ for $k \geq 100$ . Then $x_{k} \to 0$ and $(x_{k})$ is bounded, indeed $∣ x_{k} ∣ \leq 100$ for every $k$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals). The initial block is arbitrary prescribed data: convergence depends only on the tail (Convergence depends only on the tail), while the bound is read off the finite block and the tail separately. This is the worked form of the tail theorem.

Facts & Assumptions

Given: The sequence $u_{k} : = (k + 1)^{- 1}$ of The sequence $1 / (k + 1)$ converges to $0$ and the sequence $x_{k} : = k + 1$ for $k < 100$ , $x_{k} : = (k + 1)^{- 1}$ for $k \geq 100$ , where $k + 1 = σ (k)$ and $100$ are read in $R$ through the canonical-natural embedding (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Canonical naturals are positive and strictly increasing).

[F2]

Tails and convergence: $(x_{k})$ converges to $x$ if and only if every tail $x^{(K)}$ converges to $x$ ; consequently, if two sequences agree from some index on, then either both converge to $x$ or neither does (Convergence depends only on the tail).

[F3]

The reference null sequence: $u_{k} = (k + 1)^{- 1}$ converges to $0$ , with every term positive (The sequence $1 / (k + 1)$ converges to $0$ ).

[F4]

Discreteness: for all naturals $m, n$ , $m < n ⟺ σ (m) \leq n$ , so there is no natural strictly between $n$ and $σ (n)$ (Discreteness: $σ (n)$ is the immediate successor, Order on the natural numbers).

[F5]

Canonical naturals: for every $n \geq 1$ the element $n \cdot 1_{R}$ is positive, and $n \mapsto n \cdot 1_{R}$ is strictly increasing on ${1, 2, 3, \dots}$ , hence preserves non-strict order there; in particular $1 \leq 100$ and every $k + 1 \geq 1$ maps to a positive real (Canonical naturals are positive and strictly increasing).

[F6]

Reciprocals against $1$ : for $r > 0$ one has $r \geq 1 ⟺ 0 < r^{- 1} \leq 1$ , and $r = 1 ⟺ r^{- 1} = 1$ (Reciprocals and order: $1 / r$ against $1$ ).

[F7]

Inverses of positives are positive: $a > 0$ implies $a^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order).

[F8]

Absolute value: $∣ u ∣ = u$ whenever $u \geq 0$ (Absolute value in an ordered field).

[F9]

Boundedness: $(z_{k})$ is bounded when there is a real $M$ with $∣ z_{k} ∣ \leq M$ for every $k \in N$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[F10]

The order on $N$ is total: for all naturals $m, n$ exactly one of $m < n$ , $m = n$ , $n < m$ holds ( $\leq$ is a linear order on $N$ ).

Verification

technique · direct

1.1given

For every $k \geq 100$ the second clause of the definition of $(x_{k})$ gives $x_{k} = (k + 1)^{- 1} = u_{k}$ , so $(x_{k})$ and $(u_{k})$ agree from index $100$ on.

2.1step 1.1F2F3

By [F3] the sequence $(u_{k})$ converges to $0$ ; by the agreement clause of [F2] the sequence $(x_{k})$ converges to $0$ as well.

2.2step 1.1F4F5F8

For $k < 100$ : discreteness [F4] gives $k + 1 = σ (k) \leq 100$ , and $k + 1 \geq 1$ , so the real $x_{k} = k + 1$ satisfies $1 \leq x_{k} \leq 100$ by [F5] and is positive; hence $∣ x_{k} ∣ = x_{k} \leq 100$ by [F8].

2.3step 1.1F5F6F7F8

For $k \geq 100$ : the real $k + 1$ satisfies $k + 1 \geq 100 \geq 1$ , so it is positive with $(k + 1)^{- 1} \leq 1$ by [F6] and $(k + 1)^{- 1} > 0$ by [F7]; since $1 \leq 100$ by [F5], transitivity gives $0 < x_{k} \leq 100$ , hence $∣ x_{k} ∣ = x_{k} \leq 100$ by [F8].

3.1step 2.1step 2.2step 2.3F9F10∎

Every natural $k$ satisfies $k < 100$ or $k \geq 100$ by totality [F10], and steps 2.2 and 2.3 give $∣ x_{k} ∣ \leq 100$ in the two cases; so $100$ is a bound in the sense of [F9] and $(x_{k})$ is bounded. With step 2.1 this shows that changing the first hundred terms leaves both convergence to $0$ and boundedness, with the explicit bound $100$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John K. Hunter, An Introduction to Real Analysis (complete author-hosted lecture notes, 307 pp.) (standard reference, not scraped)
Jiri Lebl, Basic Analysis I: Introduction to Real Analysis, Volume I (author-hosted complete book PDF) (standard reference, not scraped)

Changing the first hundred terms leaves convergence to 00 and boundedness intact