^G,l = l 



( 311 ) 

 (7-6,1 - 7V,r,) ^Yy) + (^c.2 - n,4) ^'(4) ^ "" ~'"' " 



1111 



there follows 



1111 



7 13 19 31 



5 J G, 2 



\ 1 1 



2 14 26 38 



T - 1 1 1 1 _ 1/3 ,^, _ 1 1 1 1 



''''- "y"n"r7"23--"^' ^^'-^- + 10+22+34+ 



46 



:+- 



2jr' 

 ^/3 



2jr 



T4,iXl'[ - )+ :/\sXi^ 



11" wc take ö =: 4, we luxx'C bj applying (/.)) 



^'4,2 = ^'2,0 = O, 



and by tlie sub.stilntioii h = 4, k=l in ((r) 



IA ^ 3 ^ 1 



4y ' " v^/ ~^ 

 hence we lind 



1111 _ 2 



T~Ï3~Ï7"^2Ï ^' 



1111 _ 2 



And by snbtraction we obtain 



^.1 = 1 



T 



4,3 — — 



a(4?w+l) f<(4m4-3)' 



4 



m=0 



4m+l 4m + 3 



a result wliicii may be compared with Leibnitz's theorem 



m=co 



1 1 



?n=0 



_4m-|-l 4m-|-3 



T 



Lastly the series 71a can be evaluated if we substitute ^=:5, 

 ^•=1 in (F) and 0=5, k=i, k=2 in (G). Thus we obtain 



.T 



li.-T 



(^'5,1 + ^'5^4 ) /o// 4 f^ltr — -{- {T.^o -f 7'ó,3 ) % 4 sin'' — - zn 

 5 5 



(T5,, -T5,4) ^/y^ + {T,.-T,-,)P 



(^'5.1 - 7'5..4 ) /^ (^^ 4- (7'ó,2 - 7'5,3 ) P(~ 



moreo\'ei' we Iuxat 



T,,x + V'0,2 + ^5,3 + ^A = - 7'5.o == 0. 

 Solving- the four equations, the residt is as follows 



2jr 1 



21* 



