319 



the contimied fraction. In order to prove this convergence for s ^ 0, 

 we change /(«) into 



1 



and applying Stikltjes' method to the new integral 



r du r' r' i . 







we obtain the continued fraction 



1 I 11 1 I 1 I 11 



I— ' + !-' + I"' + I-' f I— ' + •••• 

 |a,z |a , |a ,2 |a\, |ajZ 



1 , 16 -^ , 



with ri'2n = -— and n' -m-i-x =^ n — rr- Now both tlie series 2n2k and 

 2n 271-f-l 1 



00 



2a'2k-\-i evidentl}', diverge, iience we infer that for z^O the new 







continued fraction necessarily converges, and by tlie way we may 



note for z = —- the rather remaritable result 

 16 



JX 11 11 11 11 



2 - 1 = i-r' + ly' + |-' + i-r' + • • • • 



^ It It \s It 



Comparing the functions f{i() and g{ii) we have 



•^-^^ = 1 + e-2^^« 



and accordingly everywhere in the range of integration 



l<^<2. 

 = 9W = 



therefore, again using Stikltjes' argument, we conclude to the 

 inequalities 



i{«'i+a',+«'s+"' + a'2-,+i) < (aj+«, + a, + ... + «2n+i) < 



< («'i + «'. + a's + • • • 4 a'2.1+1). 

 otherwise wi-itten 



/111 1 



< 16 7 + - + -+... +: 



13 5 2n + l 



