Mb WARBURTON, ON SELF-REPEATING SERIES. 475 



Case 1, proved. Let the repeating equation of the series be (according to equations 

 1 and 7), 



C=(-1)C ; . . . . . (12) 



x - [x + b - a] 



and let the equation of recurrence of the denominator be, 



B = (- 1) B. . . (See equation 1 1) (13) 



x b — x 



Substitute in equation (8) for B and C their respective values, as deduced from the two 

 preceding equations. Equation (8) will then become, 



A = (- + l) (BxC + B xC+ + B.C j; . . (u) 



* \a— x a + l-x 1 a — x) 



+ y + l 



that is to say, A = (- 1) A . . . . . . (15) 



x a — x 



But equation (15) is the equation of recurrence of the numerator of the fraction which 

 has been developed ; and from that equation it appears that the sign of recurrence is 



o + y + l 

 (-!)=(-!) 06) 



Consequently, when /3 and 7 are either both odd, or both even, the numerator will have 

 a negative sign of recurrence ; but when one of the said two exponents is odd, and the other 

 even, the numerator will have a positive sign of recurrence. 



Case 2, proved. As in the preceding case, let the repeating equation of the series be, 



C={-\)C ; . . . . (12) 



x - [x + 6 - a] 



and let the equation of recurrence of the numerator be, 



A = (-1) A (11) 



* a — x 



a 



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



x a — x 



substitute everywhere for Cits values, as deduced from equation (12). Equation (9) will then 

 become, 



A = (-l) + ( J BxC+fi xC+ +5xCl. . (17) 



x (&-# b\\ — x 1 b x) 



But A= BxC + B xC+ +B*C. . (5) 



x x x-1 1 Ox 



In equations (17) and (5), let 0, 1, 2, 3, &c. be substituted, in succession, for x. 



o + y + 1 



Then B = (- 1) B, 



b 



a+y + l 



Consequently, B = (- l) B . 



1 6-1 



