1.2. Stirling’s formula 37
so we heuristically have
exp(n log(1 +
s
n
) s)
exp(−s2/2n).
To achieve this approximation rigorously, we first scale s by

n to remove
the n in the denominator. Making the substitution s =

nx, we obtain
n! =

nnne−n



n
exp(n log(1 +
x

n
)

nx) dx,
thus extracting the factor of

n that we know from (1.47) has to be there.
Now, Taylor expansion tells us that for fixed x, we have the pointwise
convergence
(1.49) exp(n log(1 +
x

n
)

nx)
exp(−x2/2)
as n ∞. To be more precise, as the function n log(1 +
x

n
) equals 0 with
derivative

n at the origin, and has second derivative
−1
(1+x/

n)2
, we see from
two applications of the fundamental theorem of calculus that
n log(1 +
x

n
)

nx =
x
0
(x y)dy
(1 + y/

n)2
.
This gives a uniform lower bound
n log(1 +
x

n
)

nx
−cx2
for some c 0 when |x|

n, and
n log(1 +
x

n
)

nx −cx

n
for |x|

n. This is enough to keep the integrands exp(n log(1+
x

n
)−

nx)
dominated by an absolutely integrable function. By (1.49) and the Lebesgue
dominated convergence theorem, we thus have



n
exp(n log(1 +
x

n
)

nx) dx

−∞
exp(−x2/2)
dx.
A classical computation (based, for instance, on computing

−∞

−∞
exp(−
(x2
+
y2)/2)
dxdy in both Cartesian and polar coordinates) shows that

−∞
exp(−x2/2)
dx =


and so we conclude Stirling’s formula
(1.50) n! = (1 + o(1))

2πnnne−n.
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