CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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 central binomial coefficient is asymptotic to 4^n divided by the square root of pi n

Statement

For $n \geq 1$ , put $a_{n} : = (\binom{2 n}{n}) / 4^{n}$ . Then

$\sqrt{π n} a_{n} ⟶ 1.$

Equivalently,

$(\binom{2 n}{n}) \sim \frac{4^{n}}{\sqrt{π n}},$

where the asymptotic notation means that the ratio of the two sides tends to $1$ .

Facts & Assumptions

Given: A natural $n \geq 1$ and the positive real $a_{n} = (\binom{2 n}{n}) / 4^{n}$ .

[L1]

For $k \leq n$ , $(\binom{n}{k}) k! (n - k)! = n!$ , so the usual factorial quotient equals the binomial coefficient ( $(\binom{n}{k}) k! (n - k)! = n!$ 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 set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ , The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

[L2]

For $n \in N$ , $W_{n} = \prod_{k = 1}^{n} (2 k)^{2} / ((2 k - 1) (2 k + 1))$ , with $W_{0} = 1$ , and $W_{n} \to π / 2$ (Wallis's product: pi over two is the limit of the finite Wallis products).

[L3]

A finite product in a monoid has empty product equal to the identity and satisfies the recursion that adjoins its last factor (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity).

[L4]

Products and quotients of convergent real sequences have the corresponding limits when the limiting denominator is nonzero (Algebra of limits: sums, scalar multiples, products and quotients).

[L5]

Every nonnegative real has a unique nonnegative square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L6]

For every $ε > 0$ there is a natural $N \geq 1$ with $1 / N < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L7]

The constant $π$ is positive (Pi as twice the smallest positive zero of cosine).

Proof

technique · direct

1.1

By [L1] and [L3], $a_{n} = \frac{(2 n)!}{(n!)^{2} 4^{n}} = \prod_{k = 1}^{n} \frac{2 k - 1}{2 k} .$

givenL1L3algebra

2.1

Comparing step 1.1 with the factors in $W_{n}$ gives $W_{n} = \frac{1}{(2 n + 1) a_{n}^{2}} .$

step 1.1L2L3algebra

3.1

By [L2], [L4], and step 2.1, $((π / 2) (2 n + 1) a_{n}^{2}) \to 1$ . Also $2 n / (2 n + 1) \to 1$ by [L6], so $π n a_{n}^{2} = \frac{2 n}{2 n + 1} \cdot \frac{π}{2} (2 n + 1) a_{n}^{2} ⟶ 1.$

step 2.1L2L4L6algebra

4.1

Let $b_{n} : = \sqrt{π n} a_{n} \geq 0$ , which is defined by [L5] and [L7]. Then $b_{n}^{2} = π n a_{n}^{2} \to 1$ by step 3.1, and $∣ b_{n} - 1 ∣ = \frac{∣ b_{n}^{2} - 1 ∣}{b_{n} + 1} \leq ∣ b_{n}^{2} - 1 ∣,$ so $b_{n} \to 1$ .

step 3.1L5L7algebra

5.1

For every $n \geq 1$ , the ratio of $(\binom{2 n}{n})$ to $4^{n} / \sqrt{π n}$ is exactly $\sqrt{π n} a_{n}$ . Thus either displayed asymptotic formulation implies the other, by the definition of asymptotic equivalence.

givenstep 4.1algebra

6.1

At $n = 0$ , $a_{0} = 1$ but the comparison term $4^{n} / \sqrt{π n}$ is undefined. The theorem starts at $n = 1$ , where every denominator in steps 1.1 to 5.1 is positive, and steps 4.1 and 5.1 prove its two equivalent formulations.

givenstep 1.1step 2.1step 3.1step 4.1step 5.1L1L7∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 120 results over 27 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

D. Galvin, Primitives and techniques of integration, section 13.2 (standard reference, not scraped)
Imperial College London, History of Mathematics, Problems VI solutions (standard reference, not scraped)