Chapter 13 Sums and Asymptotics524
Theorem 13.5.1 can be proved by induction (with some pain), and there are lots
of proofs using elementary calculus, but we won’t go into them.
There are several important things to notice about Stirling’s Formula. First,.n/
is always positive. This means that
nŠ >p
2nn
en
(13.24)for alln 2 NC.
Second,.n/tends to zero asngets large. This means that
nŠp
2nn
en
(13.25)which is impressive. After all, who would expect bothandeto show up in a
closed-form expression that is asymptotically equal tonŠ?
Third,.n/is small even for small values ofn. This means that Stirling’s For-
mula provides good approximations fornŠfor most all values ofn. For example, if
we use p
2n
n
enas the approximation fornŠ, as many people do, we are guaranteed to be within a
factor of
e.n/e
1
12nof the correct value. Forn 10 , this means we will be within 1% of the correct
value. Forn 100 , the error will be less than 0.1%.
If we need an even closer approximation fornŠ, then we could use either
p
2n
n
en
e1=12nor p
2n
n
en
e1=.12nC1/depending on whether we want an upper, or a lower, bound. By Theorem 13.5.1,
we know that both bounds will be within a factor of
e
12n^1 12n^1 C 1
De1
144n^2 C12nof the correct value. Forn 10 , this means that either bound will be within 0.01%
of the correct value. Forn 100 , the error will be less than 0.0001%.
For quick future reference, these facts are summarized in Corollary 13.5.2 and
Table 13.1.