LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-01

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For fixed $k$ , $(\binom{n}{k}) / n^{k}$ tends to $1 / k!$

Statement

For each fixed $k \in N$ , $\frac{ι (\binom{n}{k})}{ι (n)^{k}} ⟶ \frac{1}{ι (k!)} (n \to \infty),$ where the expression is read for $n \geq \max {1, k}$ . For every such $n$ , one also has the uniform bound $0 \leq \frac{ι (\binom{n}{k})}{ι (n)^{k}} \leq \frac{1}{ι (k!)} .$

Facts & Assumptions

Given: A fixed $k \in N$ .

[L1]

For $n \geq k$ , $(\binom{n}{k}) k! = n^{\underline{k}}$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ , The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ), and the canonical embedding preserves the products involved (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L2]

Finite products and sums obey Finite sums and finite products, by recursion and Laws of finite sums and finite products, limits obey Algebra of limits: sums, scalar multiples, products and quotients, and $1 / ι (n) \to 0$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ). Canonical naturals are positive and strictly increasing (Canonical naturals are positive and strictly increasing), while multiplication by a positive real preserves order (Sign rules for products and monotonicity of multiplication, Ordered field).

Proof

technique · direct

1.1L1givenalgebra

For $n \geq \max {1, k}$ , $ι (\binom{n}{k}) / ι (n)^{k} = ι (k!)^{- 1} \prod_{j < k} (1 - ι (j) / ι (n))$ .

2.1step 1.1L2

For $j < k \leq n$ , strict increase and positivity give $0 \leq ι (j) / ι (n) < 1$ , so every factor in step 1.1 lies in $(0, 1]$ . Thus the finite product lies in $[0, 1]$ , proving the displayed uniform bound.

3.1step 1.1L2∎

For each of the finitely many $j < k$ , $ι (j) / ι (n) \to 0$ ; finite-product limit algebra makes the product in step 1.1 tend to $1$ . Multiplication by the fixed factor $1 / ι (k!)$ yields the limit.

For fixed $k$ , $(\binom{n}{k}) / n^{k}$ tends to $1 / k!$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For fixed kk, (nk)/nk(kn​)/nk tends to 1/k!1/k!

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For fixed $k$ , $(\binom{n}{k}) / n^{k}$ tends to $1 / k!$