In Upper Limit on the Central Binomial Coefficient, Noam Elkies and David Speyer have given a nice proof that the central binomial coefficient $\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}$. This can be used to derive Stirling's formula $$ a_n = \frac{n!e^n}{n^n \sqrt{n}} \sim \sqrt{2\pi} $$ by showing that $a_n$ is decreasing, hence convergent to some positive real number $c$, and computing $c$ via $$ c = \lim_{n \to \infty} \frac{a_n^2}{a_{2n}} = \sqrt{2\pi}. $$ My question is whether we can do without this roundabout tour and prove Stirling's formula directly along the lines of the proofs quoted above.
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asked Dec 31, 2020 at 18:53
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12$\begingroup$ Why is that more slick than the Laplace/stationary phase method for the Gamma function integral which also easily gives the complete asymptotic expansion, e.g., connecting to semiclassical expansion in terms of Feynman diagrams graded by number of loops? But +1 anyway $\endgroup$– Abdelmalek AbdesselamDec 31, 2020 at 19:58
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10$\begingroup$ Terence Tao explains the Laplace's method argument here: terrytao.wordpress.com/2010/01/02/… $\endgroup$– Qiaochu YuanDec 31, 2020 at 20:54
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2$\begingroup$ For what it is worth, that asymptotic estimate in the form $\binom{2n}{n} \sim c4^n/\sqrt{2n}$ for an unknown constant $c \approx .797$ was essentially the context in which Stirling’s formula was discovered; Stirling determined that $c = \sqrt{2/\pi}$. $\endgroup$– KConradJan 2, 2021 at 9:09
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2$\begingroup$ I think, this two-steps argument which you refer to is natural. The limit of $a_{2n}/a_n^2$ (given by central binomials asymptotics) does not change when we multiply the sequence $a_n$ by any exponential function $A^n$. That's why it is strictly weaker than the asymptotics of $a_n$. So, it is natural that two "independent" constants $\pi$ and $e$ in Stirling formula must appear "independently". $\endgroup$– Fedor PetrovJan 2, 2021 at 10:03
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1$\begingroup$ @JoséHdz.Stgo. see the third paragraph of Section 5 of kconrad.math.uconn.edu/blurbs/analysis/stirling.pdf. $\endgroup$– KConradJan 20, 2021 at 7:45
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2 Answers
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The proof in the OP based on the sequence $a_n$ is proof number 1 in Steve Dunbar's Dozen Proofs of Stirling’s Formula (page 8, worked out here). Is there an alternative proof based on a sequence $b_n$ that does not require knowledge of Stirling's formula to construct the sequence?
Proof number 2 in that list gives such an alternative, based on $b_n=\sum_{k=1}^n\log k=\log n!$ Approximation of the sum by the integral gives $$b_n\approx \int_0^n\log x\,dx+\text{error}=n\log n -n+\text{error}.$$ The calculation of the error term, to obtain the asymptotic $\log\sqrt{2\pi n}$, can be done using the Euler-MacLaurin formula.$^\ast$ If we prefer not to use Euler-MacLaurin, proofs number 3 and 4 start from the same sequence and then approximate the sum by an integral using the trapezoidal rule or a Taylor series. These calculations are worked out in Dunbar's notes.
$^\ast$ Dunbar credits this proof to John Todd, Introduction to the Constructive Theory of Functions (1963).
answered Jan 2, 2021 at 10:11
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A very elementary proof of Stirling's formula is found here or the article by M. R. Murty and K. Sampath, "A very simple proof of Stirling's formula", The Mathematics Student, Vol. 84, Nos. 1-2, January-June (2015), 129–133.
This link outlines how this proof can be done essentially in three elementary steps, with the additional assumption that we know Wallis product formula for $\pi$ (that can be proved easily by integrating $\sin ^n (x)$, see for example the Wikipedia entry). I believe this to be the most elementary proof found anywhere. It requires nothing beyond elementary Calculus.
