Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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.

The binomial double series used to re-expand a power series at an interior point is absolutely convergent and may be regrouped

Statement

Let ∑n≥0an(x−c)n\sum_{n\ge0}a_n(x-c)^n have radius RR, let dd satisfy ∣d−c∣<R|d-c|<R, and let hh satisfy ∣d−c∣+∣h∣<R|d-c|+|h|<R. Then

∑n=0∞∑k=0nι ⁣(nk)∣an∣ ∣d−c∣n−k∣h∣k<∞.\sum_{n=0}^{\infty}\sum_{k=0}^{n}\iota\!\binom nk |a_n|\,|d-c|^{n-k}|h|^k<\infty.

Consequently the binomial double series is absolutely convergent and may be regrouped by powers of hh:

∑n=0∞an(d+h−c)n=∑k=0∞(∑n=k∞ι ⁣(nk)an(d−c)n−k)hk.\sum_{n=0}^{\infty}a_n(d+h-c)^n=\sum_{k=0}^{\infty}\left(\sum_{n=k}^{\infty}\iota\!\binom nk a_n(d-c)^{n-k}\right)h^k.

Facts & Assumptions

Given: The power series and points c,d,d+hc,d,d+h from the statement.

[L1]

A power series converges absolutely at every distance smaller than its radius (A real power series converges absolutely inside its radius and diverges outside it, while either behaviour may occur at an endpoint).

[L2]

The binomial theorem gives ∑k=0nι ⁣(nk)un−kvk=(u+v)n\sum_{k=0}^{n}\iota\!\binom nk u^{n-k}v^k=(u+v)^n for all real u,vu,v (The binomial theorem in R\mathbb{R}: (x+y)n=∑k<n+1ι ⁣(nk) xky n−k(x+y)^{n} = \sum_{k<n+1} \iota\!\binom{n}{k}\, x^{k} y^{\,n-k}).

[L3]

An absolutely convergent double series may be summed and regrouped in either order (Fubini for double series: if ∑i∑j∣aij∣\sum_i \sum_j |a_{ij}| converges then both iterated sums and the sum along every bijection N→N×N\mathbb{N} \to \mathbb{N} \times \mathbb{N} converge to one and the same value, Absolutely convergent and conditionally convergent series, and the general starting index).

Proof

technique · direct
1.1

Put ρ:=∣d−c∣+∣h∣<R\rho:=|d-c|+|h|<R. By [L2], the sum of the absolute values in row nn is ∣an∣ρn|a_n|\rho^n.

givenL2algebra
2.1

The series ∑n∣an∣ρn\sum_n|a_n|\rho^n converges by [L1], so the triangular double series is absolutely convergent.

step 1.1L1
3.1

Apply the binomial theorem before summing and [L3] to regroup the absolutely convergent double series by kk. This yields the displayed identity.

step 2.1L2L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 119 results over 26 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