12 1. Preliminary Notions
which can easily be proved by observing that
(1.13)
log(n!) = log 2 + · · · + log n
2
1
log t dt +
3
2
log t dt + · · · +
n
n−1
log t dt
=
n
1
log t dt = t log t t
t=n
t=1
= n log n n + 1
log
n
e
n
.
We now state Stirling’s formula in its traditional form.
Theorem 1.14. (Stirling’s formula) As n ∞,
(1.14) n!
nne−n

2πn.
Proof. First observe that since the function log t is increasing for t 0, it
is clear that
j
j−1
log t dt log j
j+1
j
log t dt.
Adding up these inequalities, for j = 1, 2,...,n, we obtain
n
0
log t dt log n!
n+1
1
log t dt,
n log n n log n! (n + 1) log(n + 1) n.
Rearranging this last expression, we obtain
n log n n log n! n log n n + log n + O(1),
suggesting that one should consider the expression
bn := n log n n +
1
2
log n
as an approximation of log n!. So, let us consider the difference
an := log n! bn = log n! n +
1
2
log n + n.
Observe that, setting α = 1/(2n + 1) and using the fact that
1
2
log
1 + α
1 α
= α +
1
3
α3
+
1
5
α5
+ · · · ,
we get that
0 an an+1 = n +
1
2
log
n + 1
n
1
=
1
α
·
1
2
· log
1 + α
1 α
1
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