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Mathlib.Analysis.SpecialFunctions.Stirling

Stirling's formula #

This file proves Stirling's formula for the factorial. It states that $n!$ grows asymptotically like $\sqrt{2\pi n}(\frac{n}{e})^n$.

Also some global bounds on the factorial function and the Stirling sequence are proved.

Proof outline #

The proof follows: https://proofwiki.org/wiki/Stirling%27s_Formula.

We proceed in two parts.

Part 1: We consider the sequence $a_n$ of fractions $\frac{n!}{\sqrt{2n}(\frac{n}{e})^n}$ and prove that this sequence converges to a real, positive number $a$. For this the two main ingredients are

Part 2: We use the fact that the series defined in part 1 converges against a real number $a$ and prove that $a = \sqrt{\pi}$. Here the main ingredient is the convergence of Wallis' product formula for π.

Part 1 #

https://proofwiki.org/wiki/Stirling%27s_Formula#Part_1

noncomputable def Stirling.stirlingSeq (n : ) :

Define stirlingSeq n as $\frac{n!}{\sqrt{2n}(\frac{n}{e})^n}$. Stirling's formula states that this sequence has limit $\sqrt(π)$.

Equations
Instances For
    theorem Stirling.log_stirlingSeq_diff_hasSum (m : ) :
    HasSum (fun (k : ) => 1 / (2 * ↑(k + 1) + 1) * ((1 / (2 * ↑(m + 1) + 1)) ^ 2) ^ (k + 1)) (Real.log (stirlingSeq (m + 1)) - Real.log (stirlingSeq (m + 2)))

    The sequence log (stirlingSeq (m + 1)) - log (stirlingSeq (m + 2)) has the series expansion ∑ 1 / (2 * (k + 1) + 1) * (1 / 2 * (m + 1) + 1)^(2 * (k + 1)).

    The sequence log ∘ stirlingSeq ∘ succ is monotone decreasing

    @[deprecated "Use `log_stirlingSeq_diff_le` instead." (since := "2026-03-16")]
    theorem Stirling.log_stirlingSeq_diff_le_geo_sum (n : ) :
    Real.log (stirlingSeq (n + 1)) - Real.log (stirlingSeq (n + 2)) (1 / (2 * ↑(n + 1) + 1)) ^ 2 / (1 - (1 / (2 * ↑(n + 1) + 1)) ^ 2)

    We have a bound for successive elements in the sequence log (stirlingSeq k).

    theorem Stirling.log_stirlingSeq_diff_le (n : ) :
    Real.log (stirlingSeq n) - Real.log (stirlingSeq (n + 1)) 1 / (12 * n * (n + 1))

    Robbins' sharp stepwise bound for the Stirling sequence: log (stirlingSeq n) - log (stirlingSeq (n+1)) ≤ 1 / (12 n (n + 1)).

    @[deprecated "Use `log_stirlingSeq_diff_le` instead." (since := "2026-03-16")]

    We have the bound log (stirlingSeq n) - log (stirlingSeq (n+1)) ≤ 1 / (4 n ^ 2).

    For any n, we have log_stirlingSeq 1 - log_stirlingSeq n ≤ 12⁻¹.

    The sequence log_stirlingSeq is bounded below for n ≥ 1.

    The sequence stirlingSeq is positive for n > 0.

    theorem Stirling.stirlingSeq'_bounded_by_pos_constant :
    ∃ (a : ), 0 < a ∀ (n : ), a stirlingSeq (n + 1)

    The sequence stirlingSeq has a positive lower bound.

    The sequence stirlingSeq ∘ succ is monotone decreasing

    The limit a of the sequence stirlingSeq satisfies 0 < a

    Part 2 #

    https://proofwiki.org/wiki/Stirling%27s_Formula#Part_2

    theorem Stirling.tendsto_self_div_two_mul_self_add_one :
    Filter.Tendsto (fun (n : ) => n / (2 * n + 1)) Filter.atTop (nhds (1 / 2))

    The sequence n / (2 * n + 1) tends to 1/2

    theorem Stirling.stirlingSeq_pow_four_div_stirlingSeq_pow_two_eq (n : ) (hn : n 0) :
    stirlingSeq n ^ 4 / stirlingSeq (2 * n) ^ 2 * (n / (2 * n + 1)) = Real.Wallis.W n

    For any n ≠ 0, we have the identity (stirlingSeq n)^4 / (stirlingSeq (2*n))^2 * (n / (2 * n + 1)) = W n, where W n is the n-th partial product of Wallis' formula for π / 2.

    Suppose the sequence stirlingSeq (defined above) has the limit a ≠ 0. Then the Wallis sequence W n has limit a^2 / 2.

    Stirling's Formula, formulated in terms of Asymptotics.IsEquivalent.

    Global bounds #

    The Stirling sequence is bounded below by √π, for all positive naturals. Note that this bound holds for all n > 0, rather than for sufficiently large n: it is effective.

    theorem Stirling.le_factorial_stirling (n : ) :
    (2 * Real.pi * n) * (n / Real.exp 1) ^ n n.factorial

    Stirling's approximation gives a lower bound for n! for all n. The left-hand side is formulated to mimic the usual informal description of the approximation. See also factorial_isEquivalent_stirling which says these are asymptotically equivalent. That statement gives an upper bound also, but requires sufficiently large n. In contrast, this one is only a lower bound, but holds for all n. See also log_stirlingSeq_diff_le for Robbins' sharp bound on successive differences in the Stirling sequence.

    theorem Stirling.le_log_factorial_stirling {n : } (hn : n 0) :
    n * Real.log n - n + Real.log n / 2 + Real.log (2 * Real.pi) / 2 Real.log n.factorial

    Stirling's approximation gives a lower bound for log n! for all positive n. The left-hand side is formulated in decreasing order in n: the higher order terms are first. This is a consequence of le_factorial_stirling, but is stated separately since the logarithmic version is sometimes more practical, and having this version eases algebraic calculations for applications. See also log_stirlingSeq_diff_le for Robbins' sharp bound of 1/(12k(k+1)) on successive differences in the Stirling sequence, which provides finer control over the convergence rate.