By applying this m times, we obtain 
Ma 
oO 
(10) S(k,n) => (1) A; Sie + m,n +i) 
v 
HAG) (Owls PA sate, ra one REIT Oe tl pO iene soo one 
where H; is the sum of the products of the fractions 1/(k + 1), 1/(k +2), 1/(k +3), 
_.....1/(K +m), takeni at a time; Hp = 1. 
The proof of (6) $1 is as follows. If the first term of the arithmetic 
progression is zero, 
Gal i (F] (di)" = d" S(k, n) 
Il Me 
0 
and this vanishes if n<k; is (—a)* (k!) if n = k; and is 
ay eae a Pee es oes +kd] ifn =k+1. 
If the first term of the arithmetic progression is a = 0, 
k k k) 
S11 G ede Sd Say (FI (c + 1)" 
i=0 i=0 
where « = a/d = 0. 
If we use the notation 
k 
eR i(k ; 
fines => C1) 4 (x + i) 
i=0 : 
expand (a + i)” by the binomial formula and reverse the order of summation, 
we obtain 
n 
(11) f(n, x, k) = = 4 ght S(k, i) 
i=0°* 
Therefore 
f(n, x, k,) = 0 when n<k, since all the summands vanish 
= S(k,k) whenn =k 
= (n) m—t oa - 
Sea Ska) 
t=k 2) 
when n>k 
In particular, when n = k + 1 
Fé + 1,2; %) = @-+ Ls ) (k + 1)S(k, k) and on putting a/d for z, 
i ofr c. b= d Stk Ele Ge d)b @-P2d) Eo oS. + (a+ kd) 
and from these follow the three conclusions* of (6) §1. 
*Chrystal: Algebra II, Sez. 9, p. 183, gives the proof of a slightly less general theorem. 
Cauchy: Exercices 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, 
