Concrete mathematics : a foundation for computer science

466 ASYMPTOTIC3
where 0 < a,,,, < 1; this will allow us to conclude from (9.85) that
S, = F(n) + C + T (n) + f Tzk(n) + Gm,nT2m+2(n) ,
k=l
(9.86)
hence the remainder will truly be between zero and the first discarded term.
A slight modification of our previous argument will patch things up perfectly.
Let us assume that
f(2m+2'(x) 3 0 and fc2m+4)(x) 3 0, as x + 0~). (9.87)
The right-hand side of (9.85) is just like the negative of the right-hand side of
Euler’s summation formula (9.67) with a = n and b = 00, as far as remainder
terms are concerned, and successive remainders are generated by induction
on m. Therefore our previous argument can be applied.
Summation 2: Harmonic numbers harmonized.
Now that we’ve learned so much from a trivial (but safe) example, we can
readily do a nontrivial one. Let us use Euler’s summation formula to derive
the approximation for H, that we have been claiming for some time.
In this case, f(x) = l/x. We already know about the integral and derivatives
of f, because of Summation 1; also f(ml(x) = O(xpmp') as x + 00.
Therefore we can immediately plug into formula (9.85):
l