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 =

√

2π

and so we conclude Stirling’s formula

(1.50) n! = (1 + o(1))

√

2πnnne−n.