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Stirlingβs formula Shatha sattar: [email protected] October 29, 2014 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 words, we have . In other Or Proof of the stirlingβs formula first take the log of n! to get Since the log function is increasing on the interval , we get For π β₯ 1 . Add the above inequalities, with = 1,2, β¦ . π , we get Though the first integral is improper, it is easy to show that in fact it is convergent. Using the antiderivative of log( π₯) ( being x log( π₯) β π₯), we get 1 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 2 From this we get 1. the sequence {ππ } is decreasing; 1 2. the sequence {ππ β 12π} is increasing. This will imply that {ππ } 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 π π = β2π . 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 3 since we are dealing with constants, we get in fact π π = β2π This completes the proof of the Stirling's formula. 4