36 1. Preparatory material
which by Taylor expansion leads to the approximation
m+1
m
log x dx =
1
2
log m +
1
2
log(m + 1) +
m
for some error
m
=
O(1/m2).
The error is absolutely convergent; by the integral test, we have
∑n
m=1
m
= C + O(1/n) for some absolute constant C :=
∑∞
m=1
m. Performing this
sum, we conclude that
n
1
log x dx =
n−1
m=1
log m +
1
2
log n + C + O(1/n)
which after some rearranging leads to the asymptotic
(1.47) n! = (1 +
O(1/n))e1−C

nnne−n,
so we see that n! actually lies roughly at the geometric mean of the two
bounds in (1.46).
This argument does not easily reveal what the constant C actually is
(though it can in principle be computed numerically to any specified level
of accuracy by this method). To find this out, we take a different tack,
interpreting the factorial via the Gamma function Γ : R R as follows.
Repeated integration by parts reveals the
identity10
(1.48) n! =

0
tne−t
dt.
So to estimate n!, it suffices to estimate the integral in (1.48). Elementary
calculus reveals that the integrand
tne−t
achieves its maximum at t = n, so
it is natural to make the substitution t = n + s, obtaining
n! =

−n
(n +
s)ne−n−s
ds
which we can simplify a little bit as
n! =
nne−n

−n
(1 +
s
n
)ne−s
ds,
pulling out the now-familiar factors of
nne−n. We combine the integrand
into a single exponential,
n! =
nne−n

−n
exp(n log(1 +
s
n
) s) ds.
From Taylor expansion we see that
n log(1 +
s
n
) = s
s2
2n
+ . . . ,
10The
right-hand side of (1.48), by definition, is Γ(n + 1).
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