Concrete mathematics : a foundation for computer science

9.3 0 MANIPULATION 445
Namely, we know a closed form for S,: It’s just H,,z+,, - H,z. And we
know a good approximation for harmonic numbers, so we just apply it twice:
Hnz+,, = ln(n2 + n) +y +
1 1
2(n2 + n) - 12(n2 + n)2
H,z = lnn2+y+& &+O($J.
+o -$ ;
( 1
Now we can pull out large terms and simplify, as we did when looking at
Stirling’s approximation. We have
ln(n2 +n) = inn’ +ln 1 + i
( >
= lnn’+J--$+&-...;
1
n2 + n
=
1
----
1
+I-...;
n2 n3 n4
1 1
-1+3-... .
(n2 +n)2 = iT n5 n6
So there’s lots of helpful cancellation, and we find
plus terms that are O(n’). A bit of arithmetic and we’re home free:
S, = n-1 - 3-2 _ inp3 + inp4 - &np5 + An+ + o(n-‘).
(9.50)
It would be nice if we could check this answer numerically, as we did
when we derived exact results in earlier chapters. Asymptotic formulas are
harder to verify; an arbitrarily large constant may be hiding in a 0 term,
so any numerical test is inconclusive. But in practice, we have no reason to
believe that an adversary is trying to trap us, so we can assume that the
unknown O-constants are reasonably small. With a pocket calculator we find
that S4 = & + & + & + & = 0.2170107; and our asymptotic estimate when
n = 4 comes to
$(1+$(-t+ $(-;+f(f +;(-& + ;+)))) = 0.2170125.
If we had made an error of, say, & in the term for ne6, a difference of h &
would have shown up in the fifth decimal place; so our asymptotic answer is
probably correct.