Journal ArticleWe present a simple new proof and a new generalization of Boole's formula n! ∑nσj=1 (-1)n-j (n j)jn (n ∈ N)

Alzer, H

Chapman, Robin

Open Research Exeter

AUSTRALASIAN JOURNAL OF COMBINATORICSVolume 59(2) (2014), Pages 333–336On Boole’s formula for factorialsHorst AlzerMorsbacher Str. 1051545 Waldbro¨lGermanyh.alzer@gmx.deRobin ChapmanMathematicsUniversity of ExeterExeter EX4 4QFU.K.R.J.Chapman@exeter.ac.ukAbstractWe present a simple new proof and a new generalization of Boole’s for-mulan! =n∑j=1(−1)n−j(nj)jn (n ∈ N).1 IntroductionThe elegant formulan! =n∑j=1(−1)n−j(nj)jn (n ∈ N) (1.1)is given in Boole’s classical book “Calculus of Finite Diﬀerences” [3, p.5, p.19]. In2005, Anglani and Barile [2] used methods from real analysis and combinatorics toprovide two proofs. An interesting extension of Boole’s identity was published in2008 by Pohoata [3]. He applied Lagrange’s interpolating polynomial theorem toestablisha0 · bn · n! =n∑j=0(−1)n−j(nj)P (a + jb) (n ∈ N),H. ALZER AND R. CHAPMAN/AUSTRALAS. J. COMBIN. 59 (2) (2014), 333–336 334where a and b are real numbers with b = 0 and P is a real polynomial of degree nwith leading coeﬃcient a0. The special case a = 0, b = 1, P (x) = xn leads to (1.1).The aim of this note is twofold. In Section 2, we present a simple new proof of (1.1),and in Section 3, we oﬀer a new generalization of (1.1).2 A new proofHere, we apply the method of induction to obtain a short and elementary proof ofBoole’s identity. We need the following well-known formulas:j(n + 1j)= (n + 1)(nj − 1), (2.1)(n + 1j)=(nj)+(nj − 1), (2.2)n+1∑j=0(−1)n+1−j(n + 1j)jν = 0 (ν = 0, 1, ..., n). (2.3)Using the formulan+2∑j=0(−1)n+2−j(n + 2j)jν = (n + 2)ν−1∑k=0(ν − 1k) n+1∑j=0(−1)n+1−j(n + 1j)jk(ν = 1, 2, ..., n + 1)it follows by induction on n that (2.3) is valid; see also [1, chapter 2.4].Proof of (1.1). If n = 1, then both sides of (1.1) are equal to 1. Next, we assume that(1.1) holds. Applying (2.1), (2.2), (2.3) with ν = n, and the induction hypothesis weobtainn+1∑j=1(−1)n+1−j(n + 1j)jn+1 = (n + 1)n+1∑j=1(−1)n+1−j(nj − 1)jn= (n + 1)n+1∑j=1(−1)n+1−j[(n + 1j)−(nj)]jn = (n + 1)[0 +n∑j=1(−1)n−j(nj)jn]= (n + 1) · n! = (n + 1)!.This reveals that (1.1) is valid if we replace n by n + 1.3 A new generalizationWe prove the followingH. ALZER AND R. CHAPMAN/AUSTRALAS. J. COMBIN. 59 (2) (2014), 333–336 335Theorem. Let(f(j))∞j=0be a sequence of complex numbers with f(0) = 1. Then,n∑j=0(−1)n−j(nj) ∑c1+···+cj=mc1,...,cj≥0f(c1) · · ·f(cj) =∑d1+···+dn=md1,...,dn≥1f(d1) · · ·f(dn) (m,n ∈ N).(3.1)Proof. We deﬁne the formal power seriesF (X) =∞∑j=0f(j)Xj.On the one hand we have(F (X)− 1)n =n∑j=0(−1)n−j(nj)F (X)j=n∑j=0(−1)n−j(nj) ∞∑c1,...,cj=0f(c1) · · ·f(cj)Xc1+···+cj=∞∑m=0Xmn∑j=0(−1)n−j(nj) ∑c1+···+cj=mc1,...,cj≥0f(c1) · · · f(cj).On the other hand we obtain(F (X)− 1)n = (∞∑k=1f(k)Xk)n=∞∑m=nXm∑d1+···+dn=md1,...,dn≥1f(d1) · · ·f(dn).Comparing the coeﬃcients of Xm gives (3.1).RemarkThe special case m = n leads ton∑j=0(−1)n−j(nj) ∑c1+···+cj=nc1,...,cj≥0f(c1) · · ·f(cj) = f(1)n. (3.2)Since ∑c1+···+cj=nc1,...,cj≥01c1! · · · cj ! =jnn!,we conclude that formula (3.2) with f(c) = 1/c! implies (1.1).H. ALZER AND R. CHAPMAN/AUSTRALAS. J. COMBIN. 59 (2) (2014), 333–336 336References[1] M. Aigner, Discrete Mathematics, Amer. Math. Soc., Providence, R.I., 2007.[2] R. Anglani and M. Barile, Two very short proofs of a combinatorial identity,Integers: Elect. J. Comb. Number Th. 5 (2005), article A18.[3] G. Boole, Calculus of Finite Diﬀerences, 4th ed., Chelsea, New York, 1957.[4] C. Pohoata, Boole’s formula as a consequence of Lagrange’s interpolating poly-nomial theorem, Integers: Elect. J. Comb. Number Th. 8 (2008), article A23.(Received 13 Nov 2013)

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