Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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: ∑k≥12−k=1\sum_{k \ge 1} 2^{-k} = 1 and ∑k≥0(−1/3)k=3/4\sum_{k \ge 0} (-1/3)^k = 3/4

Example

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

∑k≥12−k  =  1,∑k≥0(−13)k  =  34.\sum_{k \ge 1} 2^{-k} \;=\; 1, \qquad \sum_{k \ge 0} \Big(-\frac{1}{3}\Big)^{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 ∑k≥02−k=1/(1−1/2)=2\sum_{k \ge 0} 2^{-k} = 1/(1 - 1/2) = 2, a series whose first term is 20=12^{0} = 1. The series above starts at k=1k = 1 and therefore omits that term, so its sum is 2−1=12 - 1 = 1, not 22. 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 1/21/2 and −1/3-1/3, and the integer powers rkr^{k} (Integer powers ama^m).

[L1]

For ∣r∣<1|r| < 1 the series ∑rk\sum r^{k} from the starting index 00 converges with sum 1/(1−r)1/(1-r) (For ∣r∣<1|r| < 1, ∑k≥0rk=1/(1−r)\sum_{k \ge 0} r^k = 1/(1-r), and for ∣r∣≥1|r| \ge 1 the series diverges).

[L2]

Absolute value: ∣1/2∣=1/2|1/2| = 1/2 and ∣−1/3∣=1/3|-1/3| = 1/3, both less than 11 (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 sNs_N plus the sum of the NN-th tail series (A series converges iff each of its tail series converges, and the sum splits as sNs_N plus the NN-th tail, Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion).

[L4]

r0=1r^{0} = 1 for every real rr (Integer powers ama^m, Laws of integer exponents).

Verification

technique · direct
1.1

Since ∣1/2∣=1/2<1|1/2| = 1/2 < 1, the series ∑k≥0(1/2)k\sum_{k \ge 0} (1/2)^{k} converges with sum 1/(1−1/2)=21/(1 - 1/2) = 2.

givenL1L2algebra
1.2

Since ∣−1/3∣=1/3<1|-1/3| = 1/3 < 1, the series ∑k≥0(−1/3)k\sum_{k \ge 0} (-1/3)^{k} converges with sum 1/(1−(−1/3))=1/(4/3)=3/41/(1 - (-1/3)) = 1/(4/3) = 3/4, which is the second claim.

givenL1L2algebra
1.3

The series ∑k≥12−k\sum_{k \ge 1} 2^{-k} is the 11-st tail series of ∑k≥0(1/2)k\sum_{k \ge 0} (1/2)^{k}, its terms being (1/2)j+1=2−(j+1)(1/2)^{j+1} = 2^{-(j+1)} for j∈Nj \in \mathbb{N}.

givenL3L4
2.1

The first partial sum of ∑k≥0(1/2)k\sum_{k \ge 0} (1/2)^{k} is s1=(1/2)0=1s_1 = (1/2)^{0} = 1, so by the splitting identity the tail sum is 2−1=12 - 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∣|r| and its proof runs through sn=(1−rn)/(1−r)s_n = (1 - r^{n})/(1-r), which is indifferent to the sign of rr.

  • 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 79 results over 25 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources