10 
EN ntldr). (2) 
The inequalities (42) show that the upper limit of s®) for n = ao 
is, indeed, at most equivalent to n°. On the other hand it is easily 
seen that 2? is exactly the power of n to which the upper limit 
of s@) for m= is equivalent, but this is of no importance for 
our purpose. 
We now construct a function p(t) by means of the so-determined 
series of coefficients (40) 
pO= Sn ae DEENSE ER 
0 
For this we may also write 
gy (t) = (l—t) E Sn = (1-1)? Ya se ide (43) 
0 
The last member of this equality shews of what order gp (t) is 
at most for ¢= 1-— ò. Since we have seen just now, that the upper 
limit of s@) for n — oo is at most equivalent to n°, the function 
eo 
. n s,(2) te 
0 
is, for ¢==1, at most of order 1: (1—d)!+?*, Hence p (tf) is for t= 1 
equal to zero, if O< 4, and for 4<.0< 1 at most of order 
1:(1—2)?*-!. Thus we have 
== On joes 
FOS Mor, 6.4: 
Since 206—1 = 9—(1—0) and 0 <1, we have in both cases 
1< 8. 
From the lemma proved in the preceding paragraph it now fol- 
lows, since the upper limit of s, for n =o is equivalent to n’, 
that the series of factorials 
nl ay 
POE reen nee oa (6) 
dwerges for Riv) <0. Hence it is not true here that this series con- 
verges for R(x) a r'(=0) and Rw) >A, and a fortiori not that a 
certain integral of the form (1), for all such values of x, might be 
expanded in such a series. Therefore Nivisun’s theorem is inexact. 
