Proc. Indian Acad. Sci. (Math. Sci.) Vol. 127, No. 4, September 2017, pp. 551–564. DOI 10.1007/s12044-017-0355-2 Some inequalities for the Bell numbers FENG QI1,2,3,∗ 1Institute of Mathematics, Henan Polytechnic University, Jiaozuo City 454010, Henan Province, China 2College of Mathematics, Inner Mongolia University for Nationalities, Tongliao City 028043, Inner Mongolia Autonomous Region, China 3Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City 300387, China *Corresponding author. E-mail: [email protected]; [email protected]; [email protected] MS received 28 February 2016; revised 21 June 2016; published online 19 August 2017 Abstract. In this paper, we present derivatives of the generating functions for the Bell numbers by induction and by the Faà di Bruno formula, recover an explicit formula in terms of the Stirling numbers of the second kind, find the (logarithmically) absolute and complete monotonicity of the generating functions, and construct some inequalities for the Bell numbers. From these inequalities, we derive the logarithmic convexity of the sequence of the Bell numbers. Keywords. Bell number determinant; product; inequality; generating function; derivative; absolutely monotonic function; completely monotonic func- tion; logarithmically absolutely monotonic function; logarithmically completely monotonic function; Stirling number of the second kind; induction; Faà di Bruno formula; logarithmic convexity. 2010 Mathematics Subject Classification. Primary: 11B73; Secondary: 26A48, 26A51, 33B10. 1. Introduction In combinatorics, the Bell numbers, usually denoted by Bn for n ∈{0}∪N, where N denotes the set of all positive integers, count the number of ways a set with n elements can be partitioned into disjoint and nonempty subsets. These numbers have been studied by mathematicians since the 19th century, and their roots go back to medieval Japan, but they are named after Eric Temple Bell, who wrote about them in the 1930s. The first few Bell numbers Bn are B0 = 1, B1 = 1, B2 = 2, B3 = 5, B4 = 15, B5 = 52, B6 = 203, B7 = 877, B8 = 4140, B9 = 21147. All of the Bell numbers Bn can be generated by ∞ x B ee = e k xk (1.1) k! k=0 © Indian Academy of Sciences 551 552 Feng Qi or, equivalently, by ∞ k −x x ee = e (−1)k B . k k! k=0 e±x We call e the generating functions of the Bell numbers Bk. For more information on the Bell numbers Bk, please refer to [2,5,9,16] and references therein. ±x In this paper, we present derivatives of the generating functions ee for the Bell numbers Bk by induction and by the Faà di Bruno formula, recover an explicit formula in terms of the Stirling numbers of the second kind S(n, k), find the (logarithmically) absolute and ±x complete monotonicity of the generating functions ee , and construct some inequalities for the Bell numbers Bk. From these inequalities, we derive the logarithmic convexity of the sequence of the Bell numbers Bk. ±x 2. Derivatives of the generating function ee ±x In this section, we elementarily present derivatives of the generating function ee by induction and by the famous Faà di Bruno formula. Although these results are not new and elementary, for completeness of this paper and for utilization in this paper later, we would like to state them in detail. x Theorem 2.1. For n ∈ N, the n-th derivative of the function ee can be computed by n ex n d e x = ee S(n, k)ekx, (2.1) dxn k=1 where k 1 − k S(n, k) = (−1)k n k! =1 for n ≥ k ≥ 1 are the Stirling numbers of the second kind. First proof. A straightforward computation yields ex 2 ex de x + d e x + = ee x , = ee x ex + 1 , dx dx2 3 ex d e x + = ee x e2x + 3ex + 1 , dx3 4 ex d e x + = ee x e3x + 6e2x + 7ex + 1 , dx4 5 ex d e x + = ee x e4x + 10e3x + 25e2x + 15ex + 1 , dx5 6 ex d e x + = ee x e5x + 15e4x + 65e3x + 90e2x + 31ex + 1 . dx6 Some inequalities for the Bell numbers 553 This helps us to conjecture the formula n ex n−1 d e ex +x kx = e an,ke , n ∈ N. dxn k=0 Based on this conjectured formula, a direct calculation gives n+1 ex n ex n−1 d e d d e d ex +x kx = = e an,ke dxn+1 dx dxn dx k=0 n−1 n−1 ex +x x kx ex +x kx = e (e + 1) an,ke + e kan,ke = = k 0 k 0 n−1 n−1 ex +x (k+1)x kx = e an,ke + (k + 1)an,ke = = k 0 k 0 n n−1 ex +x kx kx = e an,k−1e + (k + 1)an,ke = = k 1 k 0 n−1 ex +x nx kx = e an,n−1e + [an,k−1 + (k + 1)an,k]e + an,0 k=1 n ex +x kx = e an+1,ke . k=0 Equating the last two lines in the above equation yields an,0 = an+1,0, an,n−1 = an+1,n, n ≥ 1(2.2) and an+1,k = an,k−1 + (k + 1)an,k, 1 ≤ k ≤ n − 1. (2.3) x x The first derivative ee = ee +x means a1,0 = 1. (2.4) Combining this with the two recursions in (2.2) leads to an,0 = an+1,n = 1, n ≥ 1. (2.5) Letting k = 1in(2.3) and considering (2.4) and (2.5)give an+1,1 = an,0 + 2an,1 = 1 + 2an,1, n ≥ 2. Using a2,1 = 1 and recurring reveal that 1 n an, = 2 − 2 , n ≥ 2. (2.6) 1 2 554 Feng Qi Taking k = 2in(2.3) and using (2.6)gives n−1 an+1,2 = an,1 + 3an,2 = 2 − 1 + 3an,2, n ≥ 3. From a3,2 = 1, the above recursion figures out that 1 n n an, = 3 − 3 × 2 + 3 , n ≥ 3. (2.7) 2 3! Taking k = 3in(2.3) and using (2.7) results in 1 n n an+ , = an, + 4an, = −3 × 2 + 3 + 3 + 4an, , n ≥ 4. 1 3 2 3 6 3 From a4,3 = 1, the above recursion means that 1 n n n an, = 4 − 4 × 3 + 6 × 2 − 4 , n ≥ 4. 3 4! Similarly, we can deduce 1 n n n n an, = 5 − 5 × 4 + 10 × 3 − 10 × 2 + 5 , n ≥ 5, 4 5! 1 n n n n n an, = 6 − 6 × 5 + 15 × 4 − 20 × 3 + 15 × 2 − 6 , n ≥ 6, 5 6! 1 n n n n n n an, = 7 − 7 × 6 + 21 × 5 − 35 × 4 + 35 × 3 − 21 × 2 + 7 , 6 7! n ≥ 7. Inductively, we can conclude that k + 1 k 1 n an,k = (−1) (k − + 1) (k + 1)! =0 k+1 1 + − k + 1 = (−1)k 1 n (k + 1)! =1 = S(n, k + 1) for n ≥ k ≥ 1. The proof of Theorem 2.1 is complete. Second proof. It is easy to verify that, when n = 1, 2, the formula (2.1) is valid. A straightforward computation gives n+1 ex −1 n ex −1 n d e d d e d x = = ee −1 S(n, k)ekx dxn+1 dx dxn dx k=1 n n x x = ee +x−1 S(n, k)ekx + ee −1 S(n, k)kekx k=1 k=1 Some inequalities for the Bell numbers 555 n n x = ee −1 S(n, k)e(k+1)x + S(n, k)kekx = = k 1 k 1 n+1 n x − = ee 1 S(n, k − 1)ekx + S(n, k)kekx k=2 k=1 n x − ( + ) = ee 1 S(n, n)e n 1 x + S(n, k − 1)ekx + S(n, 1)ex k=2 n + S(n, k)kekx k=2 n x − ( + ) = ee 1 S(n, n)e n 1 x + [S(n, k − 1) + kS(n, k)]ekx + S(n, 1)ex k=2 n x − ( + ) = ee 1 S(n + 1, n + 1)e n 1 x + S(n + 1, k)ekx + S(n + 1, 1)ex k=2 n+1 x − = ee 1 S(n + 1, k)ekx, k=1 where the classical recurrence S(n + 1, k) = S(n, k − 1) + kS(n, k), 1 ≤ k ≤ n, (2.8) listed in [1, p. 825], was used in the above argument. By induction, Theorem 2.1 is proved. Third proof. In combinatorics, the Bell polynomials of the second kind, or say, the partial Bell polynomials, denoted by Bn,k(x1, x2,...,xn−k+1), are defined by − + n k 1 n! xi i B , (x , x ,...,x − + ) = n k 1 2 n k 1 n−k+1 ! = i ! i 1≤ i≤n,i ∈{0}∪N i 1 i=1 n = i=1 i i n n = i=1 i k for n ≥ k ≥ 0, see [7, p. 134, Theorem A], and satisfy 2 n−k+1 k n Bn,k abx1, ab x2,...,ab xn−k+1 = a b Bn,k(x1, x2,...,xn−k+1) and Bn,k(1, 1,...,1) = S(n, k), see [7, p. 135], where a and b are any complex numbers. The well-known Faà di Bruno formula can be described in terms of the Bell polynomials of the second kind Bn,k by n n d (k) (n−k+1) f ◦ g(x) = f (g(x))Bn,k(g (x), g (x),...,g (x)), (2.9) dxn k=0 556 Feng Qi see [7, p. 139, Theorem C].