LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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.

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Repeated poles expand formally as $(1 - λ x)^{- j} = \sum_{n \geq 0} (\binom{n + j - 1}{j - 1}) λ^{n} x^{n}$

Statement

Let $R$ be a commutative ring, let $λ \in R$ , and let $j \geq 1$ . In $R ⟦ x ⟧$ ,

$\frac{1}{(1 - λ x)^{j}} = \sum_{n \geq 0} (\binom{n + j - 1}{j - 1}) λ^{n} x^{n} .$

The binomial coefficient acts by repeated addition in $R$ . The formula includes $n = 0$ , $j = 1$ , and $λ = 0$ and is purely formal.

Facts & Assumptions

Given: A commutative ring $R$ , an element $λ \in R$ , and an integer $j \geq 1$ .

[L1]

A formal series is invertible exactly when its constant coefficient is a unit, and its inverse is unique (A formal power series is a unit exactly when its constant coefficient is a unit).

[L2]

The coefficient of a Cauchy product is $[x^{n}] (f g) = \sum_{i = 0}^{n} [x^{i}] f [x^{n - i}] g$ (Coefficient extraction is $R$ -linear, separates formal series, shifts under multiplication by $x^{k}$ , and converts products to finite convolution).

[L3]

Binomial coefficients count finite subsets and satisfy $(\binom{n}{0}) = 1$ , $(\binom{n}{n}) = 1$ , and $(\binom{n}{k}) = 0$ for $k > n$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L4]

The hockey-stick identity is $\sum_{i < n + 1} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ (Pascal's rule $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , and the hockey-stick identity $\sum_{i \leq n} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ ).

Proof

technique · induction

1.1baseL1L2L3

For $j = 1$ , put $G = \sum_{n \geq 0} λ^{n} x^{n}$ . By [L2], the constant coefficient of $(1 - λ x) G$ is $1$ and every positive coefficient is $λ^{n} - λ λ^{n - 1} = 0$ , so $G = (1 - λ x)^{- 1}$ by [L1]; this is the formula because $(\binom{n}{0}) = 1$ .

2.1ihstep 1.1L2

Assume the formula holds for one $j \geq 1$ . Multiplying its right-hand side by the $j = 1$ series from step 1.1, [L2] makes the coefficient of $x^{n}$ equal to $λ^{n} \sum_{k = 0}^{n} (\binom{k + j - 1}{j - 1})$ .

3.1step 2.1L3L4

Terms below $j - 1$ vanish by [L3], so [L4] changes the sum in step 2.1 to $(\binom{n + j}{j})$ ; therefore the product is the claimed series for exponent $j + 1$ .

4.1step 1.1step 3.1discharge-induction∎

The base case and induction step prove the formula for every $j \geq 1$ . At $n = 0$ the coefficient is $1$ , and at $λ = 0$ all positive coefficients vanish, so the stated boundaries are included.

Depends on

Used by

Dependency tree · two levels

21 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

R. P. Stanley, Enumerative Combinatorics, vol. 1, 2nd ed., Theorem 4.1.1 (standard reference, not scraped)
B. E. Sagan, Combinatorics: The Art of Counting, Section 3.7 (standard reference, not scraped)

Repeated poles expand formally as (1−λx)−j=∑n≥0(n+j−1j−1)λnxn(1−λx)−j=∑n≥0​(j−1n+j−1​)λnxn