Let g(x)=xλ=exp(λlogx), of order +∞ relative to eλ. Taking p=2 one has
(n−1)log(n−1)=(n−1)logn−1+2n1+o(n1)
whence (V, p. 225)
(n−1)n−1=e1nn−1+2e1nn−2+o1(nn−2)
and similarly
(n−2)n−2=e21nn−2+o2(nn−2);
consequently
sn=nn+e1nn−1+(2e1+e21)nn−2+o3(nn−2).
One proceeds similarly (for rn) when g is of order −∞ relative to eλ.
Now if g is of finite order μ relative to ex, and if, for example, ∑n=1∞g(n)=+∞, one can write
sn=1−e−μμ∫1ng(t)dt+m=1∑nf1(m) where
f1(n)=g(n)−1−e−μμ∫1ng(t)dt≪g(n) by prop. 6 of V, p. 239. If one has a principal part
cg1(n) of
f1(n) relative to
E′, and if one can again apply prop. 6 to the function
g1 one will obtain a primitive equivalent to
∑m=1nf1(m) if
∑n=1∞g1(n)=+∞, and equivalent to
∑m=n+1∞f1(m) in the opposite case (in the latter case one writes
∑m=1nf1(m)=C−∑m=n+1∞f1(m), with
C=∑n=1∞f1(n)).
Step by step one can thus eventually obtain an expression for sn as the sum of a certain number of primitives each of which is negligible with respect to the previous, of a term remaining negligible relative to the last primitive written, and finally a constant (the case where the remainder term tends to 0). It then remains to expand each of the primitives obtained with respect to E′ (cf. V, p. 235).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.