410 Professor Hamitton on Conjugate Functions, 
F,, (a+) by = y, but falls short of the number F,,,(a+), that is of the following 
number 
oa ~ 14. (¢+8)' , (+B)? (actBy Ga 
lt) ee ae pene COREE (101.) 
so that 
Ly=C;,, (a) x F,(B)) —Fn (a +B), (102.) 
and 
Ly< hk, (a+B)—F, (a+): (103.) 
if then we choose a positive integer 7, so as to satisfy the condition 
a 
+ 
eed (104. } 
nm+1>2(a+)5), that is 
and take m >, we shall have 
(a +f)” < 1 (a +)" , and therefore < 8, (105.) 
1x2x3x...m ~ 2-2" TxX2xX3x...2n 
however small the positive number 8 may be, and however large a + B may be, if we 
take m large enough ; but also 
Fo,(a+) —F,,(a+B)= Sey | and therefore <éxn, (106.) 
in which 
at Gae ge)! (a +)” (107.) 
~ mxl (m+i)(m+2) igs Gat 1) (m+2) x ...(2m)” 
and, therefore, 
nigh, (108.) 
because 
a+p 1 (a+f) 1 (a+)” 1 F 
m+1 2 (m+1) (m+2) Sepa es (m+1) (m+2) x ...(2 m) S om} (109.) 
therefore, combining the inequalities (103.) (106.) (108.), we find finally 
Sy <no: (110.) 
And hence, by (99.), the two sums = ¢,, © @, may both be made to approach as near 
as we desire to 0, by taking m sufficiently large; so that, in the notation of limits 
already employed, 
LZy=0, £ 3¢,=0, & 36,=0, (111.) 
