ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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.

Geometric sums computed: $\sum_{k \geq 1} 2^{- k} = 1$ and $\sum_{k \geq 0} (- 1 / 3)^{k} = 3 / 4$

Example

Two geometric sums, computed from For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges and stated with the starting index made explicit:

$\sum_{k \geq 1} 2^{- k} = 1, \sum_{k \geq 0} (- \frac{1}{3})^{k} = \frac{3}{4} .$

Both series converge, so both symbols denote (Series, partial sums, convergence and the sum, divergence, and the tail series).

The first is the one that is easy to get wrong. The theorem gives $\sum_{k \geq 0} 2^{- k} = 1 / (1 - 1 / 2) = 2$ , a series whose first term is $2^{0} = 1$ . The series above starts at $k = 1$ and therefore omits that term, so its sum is $2 - 1 = 1$ , not $2$ . A geometric series is not determined by its ratio alone; the starting index has to be said, and here it is.

Facts & Assumptions

Given: The real numbers $1 / 2$ and $- 1 / 3$ , and the integer powers $r^{k}$ (Integer powers $a^{m}$ ).

[L1]

For $∣ r ∣ < 1$ the series $\sum r^{k}$ from the starting index $0$ converges with sum $1 / (1 - r)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L2]

Absolute value: $∣ 1 / 2 ∣ = 1 / 2$ and $∣ - 1 / 3 ∣ = 1 / 3$ , both less than $1$ (Basic properties of the absolute value).

[L3]

A series converges if and only if each of its tail series converges, and then the sum splits as $s_{N}$ plus the sum of the $N$ -th tail series (A series converges iff each of its tail series converges, and the sum splits as $s_{N}$ plus the $N$ -th tail, Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L4]

$r^{0} = 1$ for every real $r$ (Integer powers $a^{m}$ , Laws of integer exponents).

Verification

technique · direct

1.1

Since $∣ 1 / 2 ∣ = 1 / 2 < 1$ , the series $\sum_{k \geq 0} (1 / 2)^{k}$ converges with sum $1 / (1 - 1 / 2) = 2$ .

givenL1L2algebra

1.2

Since $∣ - 1 / 3 ∣ = 1 / 3 < 1$ , the series $\sum_{k \geq 0} (- 1 / 3)^{k}$ converges with sum $1 / (1 - (- 1 / 3)) = 1 / (4 / 3) = 3 / 4$ , which is the second claim.

givenL1L2algebra

1.3

The series $\sum_{k \geq 1} 2^{- k}$ is the $1$ -st tail series of $\sum_{k \geq 0} (1 / 2)^{k}$ , its terms being $(1 / 2)^{j + 1} = 2^{- (j + 1)}$ for $j \in N$ .

givenL3L4

2.1

The first partial sum of $\sum_{k \geq 0} (1 / 2)^{k}$ is $s_{1} = (1 / 2)^{0} = 1$ , so by the splitting identity the tail sum is $2 - 1 = 1$ , which is the first claim.

step 1.1step 1.3L3L4∎

Remarks

The two computations use the theorem in different regimes of sign. The first has a positive ratio and a monotone sequence of partial sums; the second has a negative ratio, so its partial sums oscillate around the limit rather than climbing to it. The theorem covers both without a case split, because its hypothesis is on $∣ r ∣$ and its proof runs through $s_{n} = (1 - r^{n}) / (1 - r)$ , which is indifferent to the sign of $r$ .
Where the starting index bites. Every application of a geometric comparison on this page and its companion begins by fixing which index the comparison series starts at, precisely because the sum changes by the omitted terms while the fact of convergence does not.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

38 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 series (Wikipedia) (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

Geometric sums computed: ∑k≥12−k=1∑k≥1​2−k=1 and ∑k≥0(−1/3)k=3/4∑k≥0​(−1/3)k=3/4