476 Mb WARBURTON, ON SELF-REPEATING SERIES. 



a + y + l 



Consequently, B = (- 1) B. 



2 6-2 



And so on ; and since the several antecedent equations of the same form are true, 



a+y+l 



Consequently, B = (- l) B (18) 



x b-x 



But this is the equation of recurrence of the denominator of the fraction ; and the sign of 

 recurrence is, 



|8 a + y + l 

 (_!)=(_!) (1 9 ) 



Therefore, when a and y are both odd or both even, the denominator will have a negative 

 sign of recurrence; but when one of those two exponents is odd, and the other even, the 

 denominator will have a positive sign of recurrence. 



Case 3. The converse Proposition proved. 



Let the equations of recurrence of the numerator and of the denominator of the fraction, 

 respectively, be, 



a 



A = (- l) A , . . . . (11) 



and 5=(-l)5 . . . . (13) 



x b—x 



a 



Substitute, in equation (9), (- 1) A for A ; and on the right side of that equation, 



x a — x 



substitute everywhere for B its values, as deduced from equation (13). Equation (9) will 

 then become, 



H + /3+1 



A = (-l) (B x C +BxC + +5xC T . (20) 



x \x -(ft-a) x-1 -(b + l-a) -(x + b-a)$ 



But A= B x C +j8xC + +B x C . (5) 



x x x — 1 1 x 



In equations (20) and (5), let 0, 1, 2, 3, &c. be substituted, in succession, for x. 



o + j3+l 



Then C = (- l) C . 



-(b-a) 

 a+p + 1 



Consequently, C = (- 1) C 



1 -(b+l-a) 

 a + p + l 



Consequently, C = (- I) C 



2 - (ft + 2 - a) 



And so on ; and since the several antecedent equations of the same form are true, 



a + /3 + l 



Consequently, C = (- 1) C . . . . . (21) 



x —(x + b-a) 



But this is the equation of repetition of the series arising from the developement of the given 



