ALTERNATIVE PROOFS OF A FORMULA FOR

ALTERNATIVE PROOFS OF A FORMULA FOR BERNOULLI NUMBERS IN TERMS OF STIRLING NUMBERS FENG QI AND BAI-NI GUO

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 second kind.

1. Introduction It is well known that Bernoulli numbers Bk for k ≥ 0 may be generated by ∞ ∞ X x x X xk x2k (1.1) = B = 1 − + B2k , |x| < 2π. k x e −1 k! 2 (2k)! k=0

k=1

In combinatorics, Stirling numbers of the second kind S(n, k) for n ≥ k ≥ 0 may be computed by   k k n 1 X ` (1.2) S(n, k) = (−1)k−` ` k! `=0

and may be generated by (1.3)

∞ X (ex − 1)k xn = S(n, k) , k! n!

k ∈ {0} ∪ N.

n=k

In [5, p. 536] and [6, p. 560], the following simple formula for computing Bernoulli numbers Bn in terms of Stirling numbers of the second kind S(n, k) was incidentally obtained. Theorem 1.1. For n ∈ {0} ∪ N, we have n X k! (1.4) Bn = (−1)k S(n, k). k+1 k=0

The aim of this paper is to provide four alternative proofs for the explicit formula (1.4). 2010 Mathematics Subject Classification. Primary 11B68, Secondary 11B73. Key words and phrases. Alternative proof; explicit formula; Bernoulli number; Stirling number of the second kind; Fa` a di Bruno formula; Bell polynomial. Please cite this article as “Feng Qi and Bai-Ni Guo, Alternative proofs of a formula for Bernoulli numbers in terms of Stirling numbers, Analysis—International mathematical journal of analysis and its applications 34 (2014), no. 3, 311–317; Available online at http: //dx.doi.org/10.1515/anly-2014-0003.” This is a corrected version of the article “Bai-Ni Guo and Feng Qi, Alternative proofs of a formula for Bernoulli numbers in terms of Stirling numbers, Analysis—International mathematical journal of analysis and its applications 34 (2014), no. 2, 187–193; Available online at http://dx.doi.org/10.1515/anly-2012-1238.” 1

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F. QI AND B.-N. GUO

Remark 1.1. In [9, p. 1128, Corollary], among other things, it was found that (1.5)

B2k =

k−1 X A2(k−i) 1 1 − − 2k 2 2k + 1 2(k − i) + 1 i=1

for k ∈ N, where Am is defined by n X

mk =

m=1

k+1 X

Am n m .

m=0

It was listed in [6, p. 559] and recovered in [8, Theorem 2.1] that  (k) m  k+1 X 1 1 k = (−1) (m − 1)!S(k + 1, m) (1.6) ex − 1 ex − 1 m=1 for k ∈ {0} ∪ N. In [8, Theorem 3.1], by the identity (1.6), it was obtained that (1.7)

B2k = 1 +

2k−1 X

S(2k + 1, m + 1)S(2k, 2k − m)
2k m

m=1

−

2k X S(2k, m)S(2k + 1, 2k − m + 1) ,
2k

2k 2k + 1 m=1

k ∈ N.

m−1

In [12, Theorem 1.4], among other things, it was presented that   k−1 X X k−i−1 2k (−1)k−1 k (−1)i+` (k − i − `)2k−1 , (1.8) B2k = 2(k−1) 2k ` 2 (2 − 1) i=0

k ∈ N.

`=0

Remark 1.2. The identities in (1.6) have been generalized and applied in [7, 13]. 2. Four alternative proofs of the formula (1.4) Considering S(0, 0) = 1, it is clear that the formula (1.4) is valid for n = 0. Further considering S(n, 0) = 0 for n ≥ 1, it is sufficient to show (2.1)

Bn =

n X k=1

(−1)k

k! S(n, k), k+1

n ∈ N.

First proof. It is listed in [1, p. 230, 5.1.32] that Z ∞ −au b e − e−bu (2.2) ln = d u. a u 0 Taking a = 1 and b = 1 + x in (2.2) yields  Z ∞ Z ∞ Z 1 ln(1 + x) 1 − e−xu −u (2.3) = e du = txu−1 d t e−u d u. x xu 0 0 1/e Replacing x by ex − 1 in (2.3) results in  Z ∞ Z 1 x uex −u−1 t d t e−u d u. (2.4) = ex − 1 0 1/e

AN EXPLICIT FORMULA FOR BERNOULLI NUMBERS

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In combinatorics, Bell polynomials of the second kind (also called partial Bell polynomials) Bn,k (x1 , x2 , . . . , xn−k+1 ) are defined by (2.5)

X

Bn,k (x1 , x2 , . . . , xn−k+1 ) =

1≤i≤n,` i ∈N Pn i`i =n Pi=1 n i=1 `i =k

n! Qn−k+1 i=1

n−k+1 Y 

`i !

i=1

x i  `i i!

for n ≥ k ≥ 1, see [4, p. 134, Theorem A]. They satisfy
(2.6) Bn,k abx1 , ab2 x2 , . . . , abn−k+1 xn−k+1 = ak bn Bn,k (x1 , xn , . . . , xn−k+1 ) and (2.7)

Bn,k (1, 1, . . . , 1) = S(n, k),

see [4, p. 135], where a and b are any complex numbers. The well-known Fa`a di Bruno formula may be described in terms of Bell polynomials of the second kind Bn,k (x1 , x2 , . . . , xn−k+1 ) by (2.8)

n X
dn f ◦ g(x) = f (k) (g(x))Bn,k g 0 (x), g 00 (x), . . . , g (n−k+1) (x) , n dx k=1

see [4, p. 139, Theorem C]. Applying in (2.8) the functions f (y) = ty and y = g(x) = uex − u − 1 gives n

x

(2.9)

X
x dn tue = (ln t)k tue Bn,k uex , uex , . . . , uex . n dx k=1

Making use of the formulas (2.6) and (2.7) in (2.9) reveals x

n

x X dn tue = tue S(n, k)uk (ln t)k ekx . n dx

(2.10)

k=1

Differentiating n times on both sides of (2.4) and considering (2.10) figure out  X   Z ∞ Z 1 n dn x kx k k uex −u−1 (2.11) = S(n, k)e u (ln t) t d t e−u d u. d xn ex − 1 0 1/e k=1

On the other hand, differentiating n times on both sides of (1.1) gives   X ∞ dn x xk−n (2.12) = Bk . n x dx e −1 (k − n)! k=n

Equating (2.11) and (2.12) and taking the limit x → 0 discover  Z ∞ Z 1 n X (ln t)k d t e−u d u Bn = S(n, k) uk t 0 1/e k=1 Z n ∞ X (−1)k = S(n, k) uk e−u d u k+1 0 =

k=1 n X k=1

(−1)k k! S(n, k). k+1

The first proof of Theorem 1.1 is complete.



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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 Z 1 ln b − ln a 1 (2.13) = d t, b−a 0 (1 − t)a + tb where a, b > 0 and a 6= b. Replacing a by 1 and b by ex yields Z 1 x 1 (2.14) = d t. x − 1)t ex − 1 1 + (e 0 Applying the functions f (y) = y1 and y = g(x) = 1 + (ex − 1)t in the formula (2.8) and simplifying by (2.6) and (2.7) give   Z 1 n   dn x 1 d = dt n 1 + (ex − 1)t d xn ex − 1 0 dx Z 1X n k! Bn,k (tex , tex , . . . , tex ) d t (−1)k = x [1 + (e − 1)t]k+1 0 k=1 Z 1 n X tk (−1)k k! = B (ex , ex , . . . , ex ) d t x k+1 n,k [1 + (e − 1)t] 0 k=1 Z 1 n X → (−1)k k! tk Bn,k (1, 1, . . . , 1) d t, x → 0 = =

k=1 n X

0

(−1)k k!S(n, k)

k=1 n X

(−1)k

k=1

Z

1

tk d t

0

k! S(n, k). k+1

On the other hand, taking the limit x → 0 in (2.12) leads to   X ∞ dn x xk−n → Bn , x → 0. = Bk n x dx e −1 (k − n)! k=n

The second proof of Theorem 1.1 is thus complete.



Third proof. Let CT [f (x)] be the coefficient of x0 in the power series expansion of f (x). Then   n n X X n! (ex − 1)k k k k! S(n, k) = (−1) CT n (−1) k+1 x k+1 k=1 k=1   ∞ 1 X (ex − 1)k = n!CT n (−1)k x k+1 k=1   x 1 ln[1 + (e − 1)] − (ex − 1) = n!CT n x ex − 1   1 x = n!CT n x x e −1 = Bn . Thus, the formula (1.4) follows.



AN EXPLICIT FORMULA FOR BERNOULLI NUMBERS

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Fourth proof. It is clear that the equation (1.1) may be rewritten as ∞

ln[1 + (ex − 1)] X xk = Bk . x e −1 k!

(2.15)

k=0

Differentiating n times on both sides of (2.15) and taking the limit x → 0 reveal   ∞ X dn ln[1 + (ex − 1)] xk−n Bn = lim = lim Bk x→0 (k − n)! x→0 d xn ex − 1 k=n (k) n  X ln(1 + u) Bn,k (ex , ex , . . . , ex ), u = ex − 1 = lim x→0 u k=1 "∞ #(k) n `−1 X X u = lim (−1)`−1 Bn,k (ex , ex , . . . , ex ) x→0 ` k=1 `=1 # " ∞ n X X (` − 1)! `−k−1 `−1 u Bn,k (ex , ex , . . . , ex ) (−1) = lim x→0 (` − k − 1)!` k=1 `=k+1 " ∞ # n X X (` − 1)! `−1 `−k−1 (−1) lim = u lim Bn,k (ex , ex , . . . , ex ) u→0 x→0 (` − k − 1)!` = =

k=1 n X

k=1 n X k=1

`=k+1

(−1)k

k! Bn,k (1, 1, . . . , 1) k+1

(−1)k

k! S(n, k). k+1

The fourth proof of Theorem 1.1 is thus complete.



Acknowledgements. The authors thank Professor Doron Zeilberger for drawing our attention to the books [5, 6] and sketching the third proof in an e-mail on October 10, 2013. Thanks to his advice, the authors find that the formula (1.4) originated from the uneasily-found literature [10] and was listed as an incidental consequence of an answer to an exercise in [5, p. 536] and [6, p. 560]. References [1] M. Abramowitz and I. A. Stegun (Eds), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Applied Mathematics Series 55, 9th printing, Washington, 1970. 2 [2] P. S. Bullen, Handbook of Means and Their Inequalities, Mathematics and its Applications, Volume 560, Kluwer Academic Publishers, Dordrecht-Boston-London, 2003. 4 [3] B. C. Carlson, The logarithmic mean, Amer. Math. Monthly 79 (1972), 615–618. 4 [4] L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, Revised and Enlarged Edition, D. Reidel Publishing Co., Dordrecht and Boston, 1974. 3 [5] R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics—A Foundation for Computer Science, Addison-Wesley Publishing Company, Advanced Book Program, Reading, MA, 1989. 1, 5 [6] R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics—A Foundation for Computer Science, 2nd ed., Addison-Wesley Publishing Company, Reading, MA, 1994. 1, 2, 5 [7] B.-N. Guo and F. Qi, Explicit formulae for computing Euler polynomials in terms of Stirling numbers of the second kind, J. Comput. Appl. Math. 272 (2014), 251–257; Available online at http://dx.doi.org/10.1016/j.cam.2014.05.018. 2

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[8] B.-N. Guo and F. Qi, Some identities and an explicit formula for Bernoulli and Stirling numbers, J. Comput. Appl. Math. 255 (2014), 568–579; Available online at http://dx.doi. org/10.1016/j.cam.2013.06.020. 2 Pn k [9] S.-L. Guo and F. Qi, Recursion formulae for m=1 m , Z. Anal. Anwendungen 18 (1999), no. 4, 1123–1130; Available online at http://dx.doi.org/10.4171/ZAA/933. 2 [10] B. F. Logan, Polynomials related to the Stirling numbers, AT&T Bell Laboratories Internal Technical Memorandum, August 10, 1987. 5 [11] E. Neuman, The weighted logarithmic mean, J. Math. Anal. Appl. 188 (1994), no. 3, 885–900; Available online at http://dx.doi.org/10.1006/jmaa.1994.1469. 4 [12] F. Qi, Explicit formulas for derivatives of tangent and cotangent and for Bernoulli and other numbers, available online at http://arxiv.org/abs/1202.1205. 2 [13] A.-M. Xu and Z.-D. Cen, Some identities involving exponential functions and Stirling numbers and applications, J. Comput. Appl. Math. 260 (2014), 201–207; Available online at http://dx.doi.org/10.1016/j.cam.2013.09.077. 2 (Qi) Institute of Mathematics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China E-mail address: [email protected], [email protected], [email protected] URL: http://qifeng618.wordpress.com (Guo) School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China E-mail address: [email protected], [email protected]