By Stirling's theorem your approximation is off by a factor of $\sqrt{n}$, (which later cancels in the fraction expressing the binomial coefficients). In confronting statistical problems we often encounter factorials of very large numbers. $\begingroup$ Stirling's formula is a pretty hefty result, so the tools involved are going to go beyond things like routine application of L'Hopital's rule, although I am sure there is a way of doing it that involves L'Hopital's rule as a step. In mathematics, Stirling's approximation (or Stirling's formula) is an approximation for factorials. Introduction of Formula In the early 18th century James Stirling proved the following formula: For some = ! It is a good approximation, leading to accurate results even for small values of n . The full approximation states that , and after the proof I challenge you to bound it from above by . The factorial N! There’s something annoying about the proof – it uses a priori knowledge about . Stirling’s Approximation Last updated; Save as PDF Page ID 2013; References; Contributors and Attributions; Stirling's approximation is named after the Scottish mathematician James Stirling (1692-1770). For example, it is used in the proof of thede Moivre-Laplace theorem, which states that thenormal distributionmay be used as an approximation to thebinomial distributionunder certain conditions. )\sim N\ln N - N + \frac{1}{2}\ln(2\pi N) \] I've seen lots of "derivations" of this, but most make a hand-wavy argument to get you to the first two terms, but only the full-blown derivation I'm going to work through will offer that third term, and also provides a means of getting additional terms. I've just scanned the link posted by jspecter and it looks good and reasonably elementary. $\endgroup$ – Giuseppe Negro Sep 30 '15 at 18:21 Add the above inequalities, with , we get Though the first integral is improper, it is easy to show that in fact it is convergent. I want a result which is the other way around - a combinatorial\probabilistic proof for Stirling's approximation. Any application? First take the log of n! 2 π n n e + − + θ1/2 /12 n n n <θ<0 1 (Set-up) Let . It begins by approximating the ratio , so we had to know Stirling’s approximation beforehand to even think about this ratio. STIRLING’S APPROXIMATION FOR LARGE FACTORIALS 2 n! The Stirling formula gives an approximation to the factorial of a large number, N À 1. In its simple form it is, N! Stirling's approximation for approximating factorials is given by the following equation. Applications of Stirling’s formula can be found in di erent parts of Probability theory. This completes our proof. … µ N e ¶N =) lnN! dN … lnN: (1) The easy-to-remember proof is in the following intuitive steps: lnN! I'm not sure if this is possible, but to convince … \[ \ln(N! to get Since the log function is increasing on the interval , we get for . (because C 0). = Z ¥ 0 xne xdx (8) This integral is the starting point for Stirling’s approximation. … N lnN ¡N =) dlnN! Stirling's approximation for approximating factorials is given by the following equation. Proof of the Stirling's Formula. The inte-grand is a bell-shaped curve which a precise shape that depends on n. The maximum value of the integrand is found from d dx xne x = nxn 1e x xne x =0 (9) x max = n (10) xne x max = nne n (11) The result is applied often in combinatorics and probability, especially in the study of random walks. It is named after James Stirling , though it was first stated by Abraham de Moivre . is a product N(N-1)(N-2)..(2)(1). Stirling S Approximation To N Derivation For Info. \[ \ln(n!

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