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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

1.1

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 $(\binom{n}{0}) = 1$ .

baseL1L2L3

2.1

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

ihstep 1.1L2

3.1

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 $j + 1$ .

step 2.1L3L4

4.1

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.

step 1.1step 3.1discharge-induction∎

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 75 results over 21 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

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