LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

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.

Polynomial factors times a geometric rate decay to zero

Statement

For every real $q \in [0, 1)$ , every $p \in {1, 2}$ , and every real $δ > 0$ , there is an integer $N \geq 1$ such that $n^{p} q^{n} < δ$ for every integer $n \geq N$ .

Statement references

The natural numbers $N$ (von Neumann): the indices $n$ and $N$ are natural numbers.

Integer powers $a^{m}$ : $q^{n}$ is the natural integer power.

For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ : if $0 \leq q < 1$ , then $q^{n}$ tends to zero.

Reddi, Kale, and Kumar, Appendix B

Facts & Assumptions

Given: $q \in [0, 1)$ , $p \in {1, 2}$ , and $δ > 0$ .

[F1]

For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ : if $0 \leq s < 1$ , then for every positive tolerance there is an index after which $s^{k}$ is below that tolerance.

[F2]

Every complete ordered field is Archimedean: for every real $z$ there is an integer $N \geq 1$ with $z < N$ .

[F3]

The natural numbers $N$ (von Neumann) and Integer powers $a^{m}$ : the powers and indices use natural-number recursion, with $q^{n + 1} = q^{n} q$ .

[F4]

Ordered field, Complete ordered field (least-upper-bound property), The Cauchy-sequence reals have the least-upper-bound property, and Inverses of positives are positive, and reciprocation reverses order: the real numbers are a totally ordered field, and positive denominators preserve inequalities when dividing.

Proof

Given: $q, p, δ$ as in the Statement.

1.1givenF4algebracases

If $q = 0$ , then $n^{p} q^{n} = 0 < δ$ for every $n \geq 1$ , so take $N = 1$ . If $q > 0$ , set $s = (q + 1) / 2$ , so $q < s < 1$ , and set $A = 2^{p} - 1 \in {1, 3}$ .

2.1givenF2F3F4step 1.1choosealgebra

For every integer $n \geq 1$ , $(1 + 1 / n)^{p} \leq 1 + A / n$ : for $p = 1$ this is equality; for $p = 2$ , expand to $1 + 2 / n + 1 / n^{2}$ and use $1 / n^{2} \leq 1 / n$ . By [F2] choose $N_{0} \geq 1$ so large that $q (1 + A / N_{0}) \leq s$ . Then for every $n \geq N_{0}$ , the ratio $((n + 1)^{p} q^{n + 1}) / (n^{p} q^{n}) = q (1 + 1 / n)^{p} \leq s$ .

3.1F1F3F4step 1.1step 2.1inductionchoosealgebra∎

Induction on $k \geq 0$ using step 2.1 gives $(N_{0} + k)^{p} q^{N_{0} + k} \leq N_{0}^{p} q^{N_{0}} s^{k}$ . By [F1] choose $K$ with $N_{0}^{p} q^{N_{0}} s^{K} < δ$ . The ratio bound in step 2.1 is at most $s < 1$ , so the sequence $n^{p} q^{n}$ is nonincreasing for $n \geq N_{0}$ ; hence every $n \geq N_{0} + K$ satisfies $n^{p} q^{n} \leq N_{0}^{p} q^{N_{0}} s^{K} < δ$ . This proves the Statement in the $q > 0$ case, and step 1.1 proves the $q = 0$ case.

Depends on

Used by

Adam can have nonzero average regret on a one-dimensional convex sequence Theorem

Dependency tree · two levels

42 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

Sashank J. Reddi, Satyen Kale, and Sanjiv Kumar, On the Convergence of Adam and Beyond (standard reference, not scraped)