LemmaStatement: 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.

For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$

Statement

Let $r \in R$ and let $r^{k}$ be the integer power (Integer powers $a^{m}$ ).

If $∣ r ∣ < 1$ then $(r^{k})$ is null, that is $r^{k} \to 0$ (Limits and Cauchy sequences of reals).
If $∣ r ∣ > 1$ then $(∣ r ∣^{k})$ diverges to $+ \infty$ (Divergence to $+ \infty$ and to $- \infty$ ).

Claim 2 is stated for $∣ r ∣^{k}$ and not for $r^{k}$ on purpose: for $r < - 1$ the terms $r^{k}$ alternate in sign and are unbounded, so they neither converge nor diverge to $+ \infty$ ; what is true of them is the statement about their absolute values.

Both claims come from Bernoulli's inequality (Bernoulli's inequality $(1 + x)^{n} \geq 1 + n x$ ) and the Archimedean property. Nothing here needs the least-upper-bound property except through Every complete ordered field is Archimedean and For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ .

Facts & Assumptions

Given: A real $r$ , with integer powers as in Integer powers $a^{m}$ ; for $n \in N$ , the symbol $n$ also denotes the canonical natural $n \cdot 1_{R}$ where it occurs in an arithmetic expression.

[L1]

Absolute value: $∣ x ∣ \geq 0$ ; $∣ x ∣ = 0$ exactly when $x = 0$ ; $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ ; and $∣ x ∣ = x$ when $x \geq 0$ , so in particular $∣ 1 ∣ = 1$ because $1 > 0$ (Basic properties of the absolute value, Absolute value in an ordered field, The multiplicative identity is positive).

[L2]

Induction principle (The principle of mathematical induction), and the recursion clauses $a^{0} = 1$ , $a^{k + 1} = a^{k} a$ defining integer powers (Integer powers $a^{m}$ ).

[L3]

Bernoulli's inequality: $(1 + x)^{n} \geq 1 + n x$ for $x \geq - 1$ and $n \in N$ (Bernoulli's inequality $(1 + x)^{n} \geq 1 + n x$ ).

[L4]

Power laws: $(a b)^{n} = a^{n} b^{n}$ , and $a^{n} \neq 0$ when $a \neq 0$ (Laws of integer exponents).

[L5]

Powers and order: $a \geq 0$ gives $a^{n} \geq 0$ and $a > 0$ gives $a^{n} > 0$ ; $1^{n} = 1$ for every $n$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L6]

Reciprocals: $a > 0$ gives $a^{- 1} > 0$ ; $0 < a < b$ gives $0 < b^{- 1} < a^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order); and $0 < t < 1$ exactly when $1 / t > 1$ (Reciprocals and order: $1 / r$ against $1$ ).

[L7]

Archimedean property: for every $x \in R$ there is a natural $n \geq 1$ with $x < n$ (Every complete ordered field is Archimedean); and for every $ε > 0$ there is a natural $N \geq 1$ with $1 / N < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L8]

Canonical naturals: $n > 0$ for $n \geq 1$ , and $m \leq n$ in $N$ gives $m \leq n$ in $R$ (Canonical naturals are positive and strictly increasing).

[L9]

Multiplying inequalities of nonnegatives: $0 \leq a \leq b$ and $0 \leq c \leq d$ give $a c \leq b d$ (Multiplying inequalities of positives).

[L10]

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

[L11]

Convergence to $0$ and divergence to $+ \infty$ for a sequence of reals; a rational test value $ε > 0$ is in particular a real one (Limits and Cauchy sequences of reals, Divergence to $+ \infty$ and to $- \infty$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Proof

technique · cases

1.1

First, $∣ r^{k} ∣ = ∣ r ∣^{k}$ for every $k \in N$ , by induction: at $k = 0$ both sides are $∣ 1 ∣ = 1$ , and if $∣ r^{k} ∣ = ∣ r ∣^{k}$ then $∣ r^{k + 1} ∣ = ∣ r^{k} r ∣ = ∣ r^{k} ∣ ∣ r ∣ = ∣ r ∣^{k} ∣ r ∣ = ∣ r ∣^{k + 1}$ .

givenL1L2

1.2

Case zero. Assume $r = 0$ .

givenassume-case zero

1.3

Case small. Assume $0 < ∣ r ∣ < 1$ .

givenassume-case small

1.4

Case large. Assume $∣ r ∣ > 1$ .

givenassume-case large

2.1

In case zero, $r^{k} = 0$ for every $k \geq 1$ : indeed $r^{1} = r^{0} r = 1 \cdot 0 = 0$ , and if $r^{k} = 0$ then $r^{k + 1} = r^{k} r = 0$ , so induction gives the claim from $k = 1$ on.

step 1.2L2

2.2

In case small, put $s : = 1 / ∣ r ∣$ , which is defined since $∣ r ∣ \neq 0$ , and $h : = s - 1$ . Then $s > 1$ and $h > 0$ .

step 1.3L1L6choose

2.3

In case large, put $h^{'} : = ∣ r ∣ - 1$ , so $h^{'} > 0$ and $∣ r ∣ = 1 + h^{'}$ .

step 1.4choose

3.1

In case zero, for every rational $ε > 0$ and every $k \geq 1$ we have $∣ r^{k} - 0 ∣ = ∣ 0 ∣ = 0 < ε$ , so $r^{k} \to 0$ and claim 1 holds.

step 2.1L1L11

3.2

In case small, $∣ r ∣^{k} s^{k} = (∣ r ∣ s)^{k} = 1^{k} = 1$ , so $∣ r ∣^{k} = 1 / s^{k}$ , and $s^{k} > 0$ .

step 2.2L4L5

3.3

In case small, Bernoulli applied to $h > 0 \geq - 1$ gives $s^{k} = (1 + h)^{k} \geq 1 + k h > k h > 0$ for every $k \geq 1$ , using $1 > 0$ and $k h > 0$ .

step 2.2L3L8L9

3.4

In case large, Bernoulli applied to $h^{'} > 0 \geq - 1$ gives $∣ r ∣^{k} = (1 + h^{'})^{k} \geq 1 + k h^{'}$ for every $k \in N$ .

step 2.3L3

3.5

In case large, let $M \in R$ be arbitrary and use [L7] to fix a natural $n \geq 1$ with $M / h^{'} < n$ ; then $M \leq n h^{'}$ , since multiplying $M / h^{'} \leq n$ by $h^{'} > 0$ preserves the inequality.

step 2.3L7L9choose

3.6

In case small, let $ε > 0$ be rational; then $ε h > 0$ , so [L7] supplies a natural $N \geq 1$ with $1 / N < ε h$ , whence $1 / (N h) \leq ε$ on multiplying by $1 / h > 0$ .

step 2.2L6L7L9choose

4.1

In case small, combining steps 3.2 and 3.3: $0 < k h < s^{k}$ gives $∣ r ∣^{k} = 1 / s^{k} < 1 / (k h)$ for every $k \geq 1$ .

step 3.2step 3.3L6

4.2

In case large, for every $k \geq n$ we have $k h^{'} \geq n h^{'} \geq M$ , so $∣ r ∣^{k} \geq 1 + k h^{'} \geq 1 + M > M$ , the last step because $1 > 0$ .

step 3.4step 3.5L1L8L9

5.1

In case small, for every $k \geq N$ we have $k h \geq N h > 0$ , hence $1 / (k h) \leq 1 / (N h) \leq ε$ , and therefore $∣ r^{k} - 0 ∣ = ∣ r^{k} ∣ = ∣ r ∣^{k} < 1 / (k h) \leq ε$ .

step 1.1step 4.1step 3.6L6L8L9

5.2

In case large, an index $n$ has been produced for an arbitrary real $M$ with $∣ r ∣^{k} > M$ for all $k \geq n$ , which is exactly divergence to $+ \infty$ : claim 2 holds.

step 4.2L11

6.1

In case small, the rational $ε > 0$ was arbitrary and the index $N$ was produced from it, so $r^{k} \to 0$ and claim 1 holds.

step 5.1L11

7.1

The hypothesis $∣ r ∣ < 1$ of claim 1 is exhausted by cases zero and small, since $∣ r ∣ \geq 0$ with $∣ r ∣ = 0$ exactly when $r = 0$ , so trichotomy leaves only $0 < ∣ r ∣ < 1$ ; the hypothesis $∣ r ∣ > 1$ of claim 2 is case large. Both claims are therefore established.

step 3.1step 5.2step 6.1L1L10cases: zero small or largecases-exhaustive∎

Remarks

The two claims are not one claim in disguise. For $∣ r ∣ > 1$ the sequence $(r^{k})$ itself has no limiting behaviour to record when $r$ is negative: its terms alternate in sign and grow, so it neither converges nor diverges to $+ \infty$ nor to $- \infty$ . Stating claim 2 for $∣ r ∣^{k}$ is what makes it true as written.
The boundary $∣ r ∣ = 1$ is excluded and is genuinely different. For $r = 1$ the sequence is constant $1$ ; for $r = - 1$ it is the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ), which is bounded and divergent (FALSE: every bounded sequence converges). So neither claim extends to $∣ r ∣ = 1$ , and the two cases at the boundary do not even agree with each other.
Where this is used. Claim 1 supplies the null sequence $c^{k}$ that makes a contractive sequence Cauchy (Every contractive sequence is Cauchy, hence converges, with error bound $∣ x - x_{k} ∣ \leq c^{k - 1} ∣ x_{2} - x_{1} ∣ / (1 - c)$ for $k \geq 1$ ) and the null sequence $(1 / 10)^{n}$ that identifies the limit of the decimal truncations of $\sqrt{2}$ (The truncated decimal approximations of $\sqrt{2}$ form a Cauchy sequence of rationals with no rational limit ↗).

Depends on

Used by

…and 11 more results.

Dependency tree · two levels

46 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

Geometric progression (Wikipedia) (standard reference, not scraped)
Bernoulli's inequality (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (Thm 3.20(b)) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.5 (Lem 6.5.2) (standard reference, not scraped)

For ∣r∣<1∣r∣<1 the sequence rkrk is null, and for ∣r∣>1∣r∣>1 the sequence ∣r∣k∣r∣k diverges to +∞+∞