437 



By applying this m times, we obtain 



m 



(10) S(k, n) = 2 (— l) 1 ' Hi S(k + m, n + i) 



i=0 



k, n = 0,1,2, •; m = 1, 2, 3, 



where Hj is the sum of the products of the fractions l/(&+ 1), l/(k +2), l/{k + 3), 



l/(k + to), taken i at a time; //(, = 1. 



The proof of (6) §1 is as follows. If the first term of the arithmetic 

 progression is zero, 



(— 1)*(*] (dt) B = d n S(k,n) 



k 



k 

 and this vanishes if n<£; is ( — d) (k!) if n = k; and is 



(— d) k (k!)[d + 2d + 3d + + kd] if n = k + 1. 



If the first term of the arithmetic progression is a =•= 0, 



2(-l)* ft] (a + dif = d n 2 (-I) 1 ' fj] (x + i)» 

 i=o v ; 1=0 



where x = a/d =■= 0. 

 If we use the notation 



/(», x, k) = 2 (-1)' fjl & + o' 



expand (x + i)" by the binomial formula and reverse the order of summation, 

 we obtain 



(11) f(n,x,k) = 2 fyl x 11 -' S(k,i) 



Therefore 



/(n, x, k,) = when ra<&, since all the summands vanish 

 = S(k, k) when n = k 



^ x n ~ { S(k, i) whenn>k 



In particular, when n = k + 1 



/(* + 1, x, A) = (x + | ) (jfc + l)S(k, k) and on putting a/d for x, 



'l t+ '/(* + 1, .r, A;) = d k S(k, k)[a + (a + d) + (a + 2d) + + (a + kd)] 



and from these follow the three conclusions* of (6) §1. 



"Chrystal: Algebra II, Se ■. 9, p. 183, gives the proof of a slightly less general theorem. 

 Cauchy: Exerc-ices de mathematiques, 1826, I, p. 49 (23), obtains as a by-product the second 

 conclusion of the theorem for the case d = — 1, and remarks that it is well known. 



