Int. J. Contemp. Math. Sciences, Vol. 8, 203, no. 8, 349-353 HIKARI Ltd, www.m-hiari.com Note About a Combinatorial Sum Laurenţiu Modan Spiru Haret University, Academy of Economic Studies Department of Mathematics Faculty of Mathematics and Computer Science, Bucharest Str. I. Ghica, nr. 3, sector, Bucharest, Romania modan laurent@yahoo.fr Copyright c 203 Laurenţiu Modan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original wor is properly cited. Abstract In this short note, we present the origin story and the manner in which it can be calculated the sum: n + Mathematics Classification. Primary: 05A05, B75. Secondary: 05A0 Keywords: approximation interval, combinatorial sum, limit, Stirling formula for n!. Preliminaries In [5], we corrected an error appeared in the text and in the solution of the problem 25, chapter X, from []. Then, we proposed its following generalization: i Study in how many ways, 2n boys and 2n + girls can be partitioned in 2 teams, having 2n pearsons each. ii Find the number of all the possibilities when, in each team, it must be at least boy.
350 Laurenţiu Modan In [7], we presented the solution for the first part i of the anterior problem, namely 4n!/[2n!] 2, but for its second part ii, we only indicated the manner in which one can find the number of these claimed possibilities, given by the sum: 2 Results +. We shall study how the relation appeared. We shall state that a choice, for boys from 2n, and for 2n girls from 2n +, in the first team containing 2n persons, can be realized in: 2n ways. For the second team having also 2n persons, the number of the possibilities is: 2n 2n + 2n. 2n 2n + 2n From the two last facts, it follows that the number of the ways, in which we are interested, is given by the sum:, 2n + namely the relation. We go on using the next sum see [2], page 3 problem 55, l, where we change the notation as follows: n p, m q and m: m 0 p 0 q m p + q m With m 2n, p 2n,q 2n + in 2, we get: 2n 4n, 2n 2n or, equivalently: 2n 2 2n 4n 2n, p,q,m N. 2 22n +. 3
Note about a combinatorial sum 35 Because in the sum 3, the terms are pairwise equal, we rewrite this relation as: 4n 2n. + 2 2n Therefore, we just obtained the value of the sum from. Remar We remember here, in lin with, that in [6], we proposed the next limit to be computed: [ 2n ] lim. n n + Ξ We want now, finding an approximation interval for this sum from. Therefore we denote: Sn 4n! 2n. 4 + 2 [2n!] 2 Firstly, we shall give an evaluation for Sn, using Stirling formula of n! see [4]: n! n n e θ/2n,θ 0,. 5 e So, we find: 4n! [2n!] 2 8πn 4n 4n e e θ /48n [ 4πn 2n ] 2n 2 24n e /2n θ 4 θ 2, e e θ 2 /24n where θ,θ 2 0,. Because θ,θ 2 0,, θ4 θ 2n 2, 2n 48n holds, so: We get: e /2n 6 e /2n θ 4 θ 2 e /2n,e /48n. 7 < and then, from 8, in 4, we obtain: e /2n 4n! [2n!] 2 <e/48n, 8 2n <Sn <e /48n 2n. 9 The relation 9 gives us an approximation interval of Sn. Its another form is: e /2n < [Sn+2n +] <e /2n. 0
352 Laurenţiu Modan Remar 2 From 0, it follows that: lim [Sn+2n +]. n Ξ Remar 3 To obtain the sum, a standard combinatorial method consists in to identify the coefficient of x 2n from the both parts of the identity: + x 2n + x 2n+ +x 4n The last result of this paper, involved in the Remar 2, can be also obtained nowing that: 2m 4 m m Ω m, where: Ω m 3 5... 2m, 2 4 6... 2m and using then two sided estimations: < Ω πm+ 2 m < πm, see [3] D.S. Mitrinović, Analytic Inequalites, Springer-Verlag, Berlin-Heidelberg- New Yor, 970, pag. 92, item 3..6. This last inequality gives us: lim Ω m m π. m Passing m 2n, we obtain again the result involved in the Remar 2 Ξ. References [] Brandiburu M., Joiţa D., Nǎstǎsescu C., Niţǎ C.: Exercises and Problems for Algebra in Romanian, Rotech Pro, Printing House, 2004, Bucharest; [2] Iaglom A.M., Iaglom I.M.: Unelementary Problems Solved Elementarily; in Romanian, Editura Tehnicǎ Printing House, 983, Bucharest. [3] D.S.Mitrinović, Analytic Inequalites, Springer-Verlag, Berlin- Heidelberg-New Yor, 970;
Note about a combinatorial sum 353 [4] Modan L.: From Stirling Formulas to some Inequalities, Octogon, Math. Magazine, v. 5, nr. 2A, 2007, Braşov, pg. 833-835; [5] Modan L.: Accuracy of the Mathematic Language Used in Combinatorics in Romanian, Recreaţii Mat. v.0, nr., 2007, Iaşi, pg. 42-43; [6] Modan L.: PP 3474, Octogon, Math. Magazine, v.6, nr.b, 2008, Braşov, pg. 500; [7] Modan L.: To Teach Combinatorics Using Selected Problems, Proc. of Conf. Models in Developing Math. Ed., September 2009, Dresden, pg. 420-422. Received: December 4, 202