831 > 
= 
am 
= 0 ET 
Ten 
3 ( 1 7a 
1 
ym 
ay | (a 
2 (log i) 
hence 
ES 
T,= > Jay 
mee \P (loge) 
. pik 
En 
(log x)’ pms Va P 
pm = t, ; 
1 
; és 
En en 1 
== 0 JD 
ic =) En 
1 
5 1 
== (<5) ° (6, (log rm) 
1 
am 
ae) 
From the inequalities 
Ee 
„2m 1 
0<7T.< EAS 51<am 
pr <a n= 
p=l 
ensues 
m 
Tiss 0 (a!) 
and now only the term 7’, is to be considered. 
