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Alternative Proofs of a Formula for Bernoulli Numbers in Terms of Stirling Numbers

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Alternative Proofs of a Formula for Bernoulli Numbers in Terms of Stirling Numbers DOI 10.1515/anly-2014-0003 Ë Analysis 2014; 34 (3):311–317 Erratum Feng Qi and Bai-Ni Guo Alternative proofs of a formula for Bernoulli numbers in terms of Stirling numbers This is a corrected version of [Analysis 34 (2014), 187–193] Abstract: In the paper, the authors provide four alternative proofs of an explicit formula for computing Bernoulli numbers in terms of Stirling numbers of the sec- ond kind. Keywords: Alternative proof, explicit formula, Bernoulli number, Stirling number of the second kind, Faá di Bruno formula, Bell polynomial MSC 2010: Primary: 11B68, secondary: 11B73 ËË Feng Qi, Bai-Ni Guo: School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, P.R. China, e-mail: [email protected], [email protected] 1 Introduction It is well known that Bernoulli numbers B for k ≥ 0 may be generated by k x ∞ xk x ∞ x2k = H B = 1 − + H B , |x| < 2ð. (1.1) ex − 1 k k! 2 2k (2k)! k=0 k=1 In combinatorics, Stirling numbers of the second kind S(n, k) for n ≥ k ≥ 0 may be computed by 1 k k S(n, k) = H(−1)k−~ ~n k! ~ ~=0 and may be generated by (ex − 1)k ∞ xn = H S(n, k) , k ∈ {0} ∪ ℕ. k! n! n=k In [5, p. 536] and [6, p. 560], the following simple formula for computing Bernoulli numbers B in terms of Stirling numbers of the second kind S(n, k) was n incidentally obtained. 312 Ë F. Qi and B.-N. Guo Theorem 1.1. For n ∈ {0} ∪ ℕ, we have n k! B = H(−1)k S(n, k). (1.2) n k + 1 k=0 The aim of this paper is to provide four alternative proofs for the explicit for- mula (1.2). 2 Four alternative proofs of the formula (1.2) Considering S(0, 0) = 1, it is clear that the formula (1.2) is valid for n = 0. Further considering S(n, 0) = 0 for n ≥ 1, it is sucient to show n k! B = H(−1)k S(n, k), n ≥ 1. n k + 1 k=1 First proof. It is listed in [1, p. 230, 5.1.32] that ∞ b e−au − e−bu ln = X du. (2.1) a u 0 Taking a = 1 and b = 1 + x in (2.1) yields ∞ ∞ 1 ln(1 + x) 1 − e−xu = X e−u du = XX txu−1 dte−u du. (2.2) x xu 0 0 1/e Replacing x by ex − 1 in (2.2) results in ∞ 1 x = XX tue −u−1 dte−u du. (2.3) ex − 1 x 0 1/e In combinatorics, Bell polynomials of the second kind (also called partial Bell polynomials) B (x , x , . , x ) are dened by n,k 1 2 n−k+1 n! n−k+1 x ~ B (x , x , . , x ) = H I i n,k 1 2 n−k+1 ∏n−k+1 ~ ! i! i 1≤i≤n,~ ∈ℕ i=1 i i=1 ∑ i~ =n i ∑n ~ =k i=1 i n i for n ≥ k ≥ 1, see [4, p. 134, Theorem A].i= They1 satisfy B abx , ab2x , . , abn−k+1x = akbnB (x , x , . , x ) (2.4) n,k 1 2 n−k+1 n,k 1 n n−k+1 An explicit formula for Bernoulli numbers Ë 313 and n−k+1 B 1, 1, . , 1 = S(n, k), (2.5) n,k see [4, p. 135], where a and b are any complex numbers. The well-known Faà di Bruno formula may be described in terms of Bell polynomials of the second kind B (x , x , . , x ) by n,k 1 2 n−k+1 dn n f ∘ g(x) = H f(k)(g(x))B gὔ(x), gὔὔ(x), . , g(n−k+1)(x), (2.6) dxn n,k k=1 see [4, p. 139, Theorem C]. Applying in (2.6) the function f(y) = ty and y = g(x) = uex − u − 1 gives dntue n n−k+1 x = H(ln t)ktue B uex, uex, . , uex . (2.7) dxn x n,k k=1 Making use of the formulas (2.4) and (2.5) in (2.7) reveals dntue n x = tue H S(n, k)uk(ln t)kekx. (2.8) dxn x k=1 Dierentiating n times on both sides of (2.3) and considering (2.8), we obtain ∞ 1 dn x n = H S(n, k)ekx X ukX(ln t)ktue −u−1 dte−u du. (2.9) dxn ex − 1 x k=1 0 1/e On the other hand, dierentiating n times on both sides of (1.1) gives dn x ∞ xk−n = H B . (2.10) dxn ex − 1 k (k − n)! k=n Equating (2.9) and (2.10) and taking the limit x → 0, we deduce ∞ 1 n (ln t)k B = H S(n, k) X ukX dte−u du n t k=1 0 1/e ∞ n (−1)k = H S(n, k) X uke−u du k + 1 k=1 0 n (−1)kk! = H S(n, k). k + 1 k=1 The rst proof of Theorem 1.1 is complete. 314 Ë F. Qi and B.-N. Guo Second proof. In the book [2, p. 386] and in the papers [3, p. 615] and [11, p. 885], it was given that 1 ln b − ln a 1 = X dt, b − a (1 − t)a + tb 0 where a, b > 0 and a ̸= b. Replacing a by 1 and b by ex yields 1 x 1 = X dt. ex − 1 1 + (ex − 1)t 0 Applying the functions f(y) = 1 and y = g(x) = 1 + (ex − 1)t in the formula (2.6) y and simplifying by (2.4) and (2.5) give 1 dn x dn 1 = X dt dxn ex − 1 dxn 1 + (ex − 1)t 0 1 n k! n−k+1 = X H(−1)k B tex, tex, . , tex dt [1 + (ex − 1)t]k+1 n,k 0 k=1 1 n tk n−k+1 = H(−1)kk! X B ex, ex, . , ex dt [1 + (ex − 1)t]k+1 n,k k=1 0 1 n n−k+1 → H(−1)kk! X tkB 1, 1, . , 1 dt, x → 0 n,k k=1 0 1 n = H(−1)kk!S(n, k) X tk dt k=1 0 n k! = H(−1)k S(n, k). k + 1 k=1 On the other hand, taking the limit x → 0 in (2.10) leads to dn x ∞ xk−n = H B → B , x → 0. dxn ex − 1 k (k − n)! n k=n The second proof of Theorem 1.1 is thus complete. An explicit formula for Bernoulli numbers Ë 315 Third proof. Let CT[f(x)] be the coecient of x0 in the power series expansion of f(x). Then n k! n n! (ex − 1)k H(−1)k S(n, k) = H(−1)kCT k + 1 xn k + 1 k=1 k=1 1 ∞ (ex − 1)k = n!CT H(−1)k xn k + 1 k=1 1 ln[1 + (ex − 1)] − (ex − 1) = n!CT xn ex − 1 1 x = n!CT xn ex − 1 = B . n Thus, the formula (1.2) follows. Fourth proof. It is clear that the equation (1.1) may be rewritten as ln[1 + (ex − 1)] ∞ xk = H B . (2.11) ex − 1 k k! k=0 Dierentiating n times on both sides of (2.11) and taking the limit x → 0 reveal ∞ xk−n B = lim H B n k x→0 (k − n)! k=n dn ln[1 + (ex − 1)] = lim x→0 dxn ex − 1 n ln(1 + u) (k) n−k+1 = lim H B ex, ex, . , ex , u = ex − 1 n,k x→0 u k=1 n ∞ u~−1 (k) n−k+1 = lim H H(−1)~−1 B ex, ex, . , ex n,k x→0 ~ k=1 ~=1 n ∞ (~ − 1)! n−k+1 = lim H H (−1)~−1 u~−k−1B ex, ex, . , ex n,k x→0 (~ − k − 1)!~ k=1 ~=k+1 n ∞ (~ − 1)! n−k+1 = H lim H (−1)~−1 u~−k−1 lim B ex, ex, . , ex n,k u→0 (~ − k − 1)!~ x→0 k=1 ~=k+1 n k! n−k+1 = H(−1)k B 1, 1, . , 1 k + 1 n,k k=1 n k! = H(−1)k S(n, k). k + 1 k=1 The fourth proof of Theorem 1.1 is thus complete. 316 Ë F. Qi and B.-N. Guo Remark 2.1. In [9, p. 1128, Corollary], among other things, it was found that 1 1 k−1 A B = − − 2k H 2(k−i) 2k 2 2k + 1 2(k − i) + 1 i=1 for k ∈ ℕ, where A is dened by m n k+1 H mk = H A nm. m m=1 m=0 It was listed in [6, p. 559] and recovered in [8, Theorem 2.1] that 1 (k) k+1 1 m = (−1)k H (m − 1)!S(k + 1, m) (2.12) ex − 1 ex − 1 m=1 for k ∈ {0} ∪ ℕ. In [8, Theorem 3.1], by the identity (2.12), it was obtained that 2k−1 S(2k + 1, m + 1)S(2k, 2k − m) B = 1 + H 2k 2k m=1 m 2k 2k S(2k, m)S(2k + 1, 2k − m + 1) − H , k ∈ ℕ.
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