Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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.

A convergent series in R2 with Γ a line and Γ⊥ a line, computed from the definition

Example

Let (εk) be the alternating sequence, the unique sequence of reals with ε0=1 and εk+1=−εk, so ∣εk∣=1 for every k (The even and odd index maps and the alternating sequence: strictly increasing e,o with N their disjoint union, and the unique (sk) with s0=1, sσ(k)=−sk, which satisfies ∣sk∣=1, s∘e≡1 and s∘o≡−1). In R2 put

xk  :=  (εkι(k+1), 0)(k∈N),

with ι the canonical natural (The canonical natural ι(n)=n⋅1F of a field). Call a line through the origin the set of scalar multiples of a fixed nonzero vector; each such set is a linear subspace (Linear subspace of a vector space). Then:

  1. ∑xk converges, to s=(S,0) where S is the sum of the alternating harmonic series; the value of S is not computed here, being a logarithm and outside this page's reach.
  2. ∑xk does not converge absolutely (Series of vectors in Rn, absolute convergence, rearrangement, and the set of rearrangement sums).
  3. Γ={ (0,t):t∈R }, the line of multiples of e1, and Γ⊥={ (t,0):t∈R }, the line of multiples of e0 (The subspace Γ of directions along which a series converges absolutely, and its orthogonal complement Γ⊥).
  4. Consequently The set of rearrangement sums of a convergent series in Rn is a nonempty subset of the affine subspace s+Γ⊥ confines every rearrangement sum to the horizontal line s+Γ⊥={(t,0):t∈R}; and for this series the confinement is exact, S(x)=s+Γ⊥, by the published The Riemann series theorem: a conditionally convergent real series has, for every c∈R, a rearrangement with sum c, and rearrangements diverging to +∞, to −∞, and oscillating with any prescribed lim inf⁡≤lim sup⁡ in R‾ applied to the first coordinate.

Clause 4 decides nothing about the general question. This series is degenerate: it lies inside a line, so its rearrangement behaviour is the one-dimensional behaviour of its first coordinate and nothing more. It is therefore not evidence about whether S(x)=s+Γ⊥ for a series genuinely spread over Rn with n≥2, a question this library does not settle (Conventions of this page, the standing n≥1 hypothesis, and what is taken up elsewhere in the reading order).

Facts & Assumptions

Given: The sequence (xk) above, its first coordinate sequence ck:=εk/ι(k+1) and the sequence bk:=1/ι(k+1).

[L4]

The p-series theorem: ∑k≥11/kp converges if and only if p>1; at p=1 the harmonic series diverges (For rational p>0, ∑1/kp converges iff p>1, Rational powers ar of a positive base, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

For c≠0, ∑c ak converges if and only if ∑ak converges (Convergent series add and scale termwise clause 3).

Verification

technique · direct
1.1

(bk) is positive, nonincreasing and converges to 0: positivity and monotonicity from 0<ι(k+1)<ι(k+2), and convergence because for a rational ε>0 an index K with 1/ι(K+1)<ε gives bk≤bK<ε for k≥K.

L2
1.2

The second coordinate sequence is constantly 0, so its series converges with sum 0.

L5
2.1

By the alternating series test ∑kck=∑kεkbk converges; write S for its sum.

step 1.1L3
2.2

∥xk∥2=ck2+0=∣ck∣=∣εk∣bk=bk, and ∑kbk is the harmonic series, which diverges; so ∑xk does not converge absolutely, which is clause 2.

step 1.1L1L4L6
3.1

By componentwise convergence, ∑xk converges with sum s=(S,0), which is clause 1.

step 2.1step 1.2L5
3.2

For a=(a0,a1)∈R2: ⟨a,xk⟩=a0ck, so ∣⟨a,xk⟩∣=∣a0∣ bk. If a0=0 every term is 0 and the series converges; if a0≠0 then ∣a0∣>0 and convergence of ∑k∣a0∣bk would give convergence of ∑kbk, which is false.

step 2.2L1L6L7
3.3

Conversely let t∈R. The real series ∑kck converges by step 2.1 and does not converge absolutely by step 2.2, so it converges conditionally, and the Riemann series theorem supplies a bijection σ of N with ∑kcσ(k)=t. The rearranged vector series ∑kxσ(k) has first coordinate series ∑kcσ(k) and second coordinate series constantly 0, so by componentwise convergence it converges to (t,0); hence (t,0)∈S(x).

step 2.1step 2.2L5L8
4.1

Hence Γ={a:a0=0}={(0,t):t∈R}, the set of scalar multiples of e1.

step 3.2L6
5.1

For y=(y0,y1): y∈Γ⊥ means ⟨(0,t),y⟩=t y1=0 for every real t, which at t=1 forces y1=0, and conversely y1=0 makes every such product 0. So Γ⊥={(t,0):t∈R}, the set of scalar multiples of e0, and clause 3 is proved.

step 4.1L6
6.1

By the containment theorem, S(x)⊆s+Γ⊥={ (S+t, 0):t∈R }={ (w,0):w∈R }.

step 3.1step 5.1L8
7.1

Steps 6.1 and 3.3 give S(x)=s+Γ⊥, which is clause 4.

step 6.1step 3.3∎

Remarks

Depends on

Used by

Dependency tree · two levels

137 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