Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-16
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The binomial double series for re-expanding a complex power series is absolutely convergent and may be regrouped

Statement

Let ∑cn(z−a)n have radius R. If ∣b−a∣+∣h∣<R, then ∑n≥0∑k=0n(nk)∣cn∣ ∣b−a∣n−k∣h∣k<∞, and the complex binomial double series may be regrouped by powers of h.

Facts & Assumptions

Given: A complex power series and points b,h satisfying ∣b−a∣+∣h∣<R.

[L1]

The corresponding nonnegative real binomial double series converges and licenses regrouping (The binomial double series used to re-expand a power series at an interior point is absolutely convergent and may be regrouped).

[L2]

For complex z,w and n∈N, (z+w)n=∑k≤n(nk)zkwn−k (The binomial theorem over the complex field).

[L3]

Every absolutely convergent complex series converges, and every rearrangement has the same sum (Every absolutely convergent complex series converges, and rearrangements preserve its sum).

Proof

technique · direct
1.1L1

Apply [L1] to the real coefficient sequence ∣cn∣ and the nonnegative numbers ∣b−a∣,∣h∣; this gives the displayed finite total majorant.

2.1step 1.1L2

By [L2], cn((b−a)+h)n is the finite sum over k≤n of the corresponding complex terms, each bounded by the majorant term in step 1.1.

3.1step 1.1step 2.1L3∎

Absolute convergence now permits regrouping by k under [L3]. The cases h=0 and b=a merely make some terms vanish and are included.

Depends on

Used by

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Sources