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xn Proof that lim 0 n n! By Stirling’s Approximation for Factorials ╬ Francis J. O’Brien, Jr., Ph.D. Aquidneck Indian Council Newport, RI November 15, 2012 Introduction xn The proof that lim 0 (all x), is given in the calculus texts by Finney and n n! Thomas (page A-14, Appendix), Bers (p. 506) as well as others. This is an important limit since factorials are used in many applications in mathematics and probability as well as science and engineering. The standard method of proof is by the Sandwich Theorem for Sequences (see Finney and Thomas, Theorem 2 and Theorem 3, pp. 580 ff.). Some students find the proof not always easy to follow for this limit. In this paper we provide an alternative elementary justification for the proof by using Stirling’s approximation formula for factorials. That is, rather than using n! directly in the calculation of the limit, we use an approximation for n! which is known to very accurate for large values of n. We get the same answer of 0 by this approach, but the method is not as elegant as the Sandwich Theorem for Sequences. However, limits which the Sandwich Theorem proves can also be done in other (usually longer, more cumbersome) ways such as l’Hôpital’s Rule for indeterminant forms by derivatives, and other methods; e.g., 1 lim 0 n 2n can be demonstrated by first taking the log of the function , or, 1 ln y ln lim ln 2 lim n , y e ln y 0. The Integral Test also works on this n 2 n n 1 function, e n ln 2 . Profs. Finney and Thomas use the Sandwich Theorem for n 2 Sequences to get the quickest (and more elegant) solution. Stirling’s Approximation for Factorials Feller (pp. 50 ff.) provides a good proof of the Stirling approximation. Boas (pp. 552 ff.) provides a good proof outline of the complicated derivation of Stirling’s asymptotic formula for n! by using the gamma function, change of variable, and series expansions. The factorial n! is first expressed by the gamma function, © Francis J. O’Brien, Jr., Aquidneck Indian Council <> All rights reserved. Page 1 of 6 n 1 n! x n e x dx. 0 The calculations of Boas lead to an approximation of n! : n n n! n n e n 2n 2n n e where we use the approximation equality . This approximation of n! is the relation used in xn practice for large values of n and this is used to show the limit, lim 0. n n! One other useful tool we need is the natural log of n! of the Stirling approximation. 1 ln n! ln n n e n 2n 2 1 ln n n ln e n ln 2n 2 n ln n n. The last term ln 2n 2 is dropped since for large n it is so small. For example, take a large 1 number like, n 10 20 . Then, by the approximation for ln n! ln 10 20 ! , ln n! 10 20 ln(10 20 ) 10 20 ln 2 10 20 4.5121 23.94. The relative contribution of the value of ln 2 10 20 23.94 is so small it is ignored. Thus, we take ln n! n ln n n in our calculations for the limit demonstration. xn To provide an intuitive justification for the infinite limiting value of , Figure 1 n! xn below shows a plot of the function y for x = 10 and x = 100 and variable n. In each n! case we see visual evidence of convergence to zero as n . Now it must be shown by proof when n! is approximated. © Francis J. O’Brien, Jr., Aquidneck Indian Council <> All rights reserved. Page 2 of 6 n n xn Figure1. Plots of y . Left Graph (x = 10); Right (x = 100)—connecting top of n! this graph not displayed. Source: Wolfram Mathematica, http://www.wolframalpha.com/. NOTE: The peak maximum y values found by differentiation using the Chain Rule. See below. Proof xn The limit to prove is y lim 0 , where n is assumed integer. For convenience, n n! we omit writing the y in each step. 1. Set numerator and denominator to exponential form by definition of x e ln x , Laws of Exponents, and Stirling’s Formula for ln n! : x n e n ln x n! e ln n ! e n ln n n n 2. Simplify by Laws of Exponents, e nx e x : xn e n ln x lim e n ln x n ln n n lim e ln x ln n 1 n lim lim n n! n e n ln n n n n 3. Expand limit n times by e x n enx e x e x : times n lim e ln x ln n 1n lim e ln x ln n 1 lim e ln x ln n 1 n n n 4. Take log of limit product, expand by Product Rule for Logs, and use the property that the log of lim is lim of log: © Francis J. O’Brien, Jr., Aquidneck Indian Council <> All rights reserved. Page 3 of 6 n n ln lim e ln x ln n 1 lim e ln x ln n 1 lim ln e ln x ln n 1 lim ln e ln x ln n 1 n n 5. Use log of exponents law, ln e x x : lim ln e ln x ln n 1 lim ln e ln x ln n 1 lim ln x ln n 1 lim ln x ln n 1 n n n n 6. Use the limit, lim ln n : n limln x ln n 1 limln x ln n 1 ln x 1 ln x 1 n n 7. Use definition of exponential of logs, y e ln y : y e 0. That is, up to this point we have worked with the functions called y & ln y . We just found ln y to be . xn 8. Concludes the proof, lim 0 by Stirling’s approximation for n! n n! en NOTE: The derivation indicates that lim 0 which may be surprising since e n grows n n! so fast, but the calculated limit apply to all x values. xn Derivative of n! By above transform to logs in Step 2, x n e n ln x e n ln x n ln n n n! e ln n! d u du Then, by the Chain Rule, e eu , with u n ln x n ln n n , and dn dn differentiating, © Francis J. O’Brien, Jr., Aquidneck Indian Council <> All rights reserved. Page 4 of 6 d xn xn d n ln x n ln n n dn n! n! dn xn ln x ln n 1 1 0 n! ln x ln n e ln x e ln n or x n. xn Thus, y is max at n x by the graph examples in Fig. 1. We know by the limit n! xn calculation that the function decreases steadily as n increases. The function y rises, n! peaks at the value of n x for a selected x, and then falls rapidly to the value of 0. The nn approximate max y value is given by . n! NOTE: See Morrey [University Calculus (p. 462)] who explains how to prove that such a function converges for all x values by the Ratio Test. nn Calculations by substituting n 10, n 100 into will give the y values on the n! graphs by Mathematica in Figure 1. For small n = 10, an exact answer follows by 1010 100100 multiplication of 10! and direct substitution, y 2756. For n =100, y is too 10! 100! big to handle for a calculator, so to get an approximate answer, nn nn e n e100 1.1x10 42 11x10 41. n n e n 2n n! 2n 200 Other Limits by Stirling’s Formula Boas (p. 554) gives a list of Problems involving factorials which can be solved by Stirling’s Formula for n! For example, lim 2n ! n . n 2 2 n n!2 1 The answer turns out to be by appropriate substitutions of n! n n e n 2n and limit rules. n! Other limits can be evaluated by Stirling, such as lim . One might think this is n n n xn n! the flip of lim 0 or lim . But it also turns out to be equal to 0 by Stirling: n n! n n n © Francis J. O’Brien, Jr., Aquidneck Indian Council <> All rights reserved. Page 5 of 6 n! e ln n! lim lim e n ln n n n ln n lim e n 0. n n n e n ln n n n xn n! The difference with lim is that here x is a fixed quantity but in lim , n is a n n! n n n moving quantity. References Bers, L. Calculus, Vol. 1, 1969. Holt, Rinehart and Winston, Inc. Boas, Mary L. Mathematical Methods in the Physical Sciences, 3rd ed., 2006. NY: Wiley. Feller, William. Introduction to the Theory of Probability and Its Applications. 2nd ed. Vol. I., NY: John Wiley and Sons. 1957. Finney, R.L. and G.B. Thomas, Jr. Calculus, 1990. Addison-Wesley Publishing Co. Morrey, C. B., Jr. University Calculus, 1962. Addison-Wesley Publishing Co. Weisstein, Eric W. "Stirling's Approximation." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/StirlingsApproximation.html Wikipedia. Stirling's approximation. http://en.wikipedia.org/wiki/Stirling%27s_approximation © Francis J. O’Brien, Jr., Aquidneck Indian Council <> All rights reserved. Page 6 of 6 6 of 6