# Stirling's Formula An important formula in applied mathematics as well as in probability is the Stirling's formula known as where is used to indicate that the ratio of the two sides goes to 1 as n goes to . In other words, we have or ## Proof of the Stirling's Formula

First take the log of n! to get Since the log function is increasing on the interval , we get for . Add the above inequalities, with , we get Though the first integral is improper, it is easy to show that in fact it is convergent. Using the antiderivative of (being ), we get Next, set We have Easy algebraic manipulation gives Using the Taylor expansion for -1 < t < 1, we get This implies We recognize a geometric series. Therefore we have From this we get
1.
the sequence is decreasing;
2.
the sequence is increasing.
This will imply that converges to a number C with and that C > d1 - 1/12 = 1 - 1/12 = 11/12. Taking the exponential of dn, we get The final step in the proof if to show that . This will be done via Wallis formula (and Wallis integrals). Indeed, recall the limit Rewriting this formula, we get Playing with the numbers, we get Using the above formula we get Easy algebra gives since we are dealing with constants, we get in fact . This completes the proof of the Stirling's formula. [Trigonometry] [Calculus]
[Geometry] [Algebra] [Differential Equations]
[Complex Variables] [Matrix Algebra] S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard. Mohamed A. Khamsi
Tue Dec 3 17:39:00 MST 1996