Derivations of Stirling’s Approximation—C.E. Mungan, Spring 1998
Method 1: By Taylor Series
Begin with
∞
∞
n −x
n! = Γ (n + 1) = ∫ x e dx = ∫ e n ln x − x dx
0
0
and for convenience define f ( x ) = n ln x − x . By graphing f(x), you can convince yourself that it
peaks at x = n since f ′(n) = 0 . Hence, we can approximate f(x) by expanding it around x = n. (By
expanding around the maximum, the integral will include most of the area.) We get
f ( x ) = f (n) + f ′(n) ⋅ ( x − n) + 12 f ′′(n) ⋅ ( x − n)2 + L
⇒ e f ( x ) = n ne − n × e −( x − n) / 2 n × L
since f ′′(n) = −1 / n . Recalling the normalization of a Gaussian of mean n and standard deviation
n , we immediately obtain
2
n −n
n!~ n e
∞
−( x − n)
∫e
2
/ 2( n ) 2
dx = n ne − n 2πn
0
which is Boas Eq. (11.11.1). I extended the lower limit in evaluating the integral to –∞ since n is
assumed to be large.
Method 2: Using Infinite Series, Products, and Limits
This method due to Mermin is longer, but enlightening. By inspection we see that
n! = 1 × 2 × 3 × L × (n − 1) × n
=
( 12 )3 / 2 × ( 23 )5 / 2 × ( 43 )7 / 2 × L × ( nn−1 )n −1 / 2 × nn +1 / 2
∞
=
n
n −1
n +1 / 2
∏ (1 + m1 )
n +1 / 2
m +1 / 2
n
= n −1
e
∏ (1 + m1 )
m=n
∞
∏ (1 + m1 )
m +1 / 2
m +1 / 2
/e
/e
m =1
m =1
Let’s define a constant P such that
∞
e
m +1 / 2
/e ≡
∏ (1 + m1 )
2P
m =1
(1a)
which is meaningful because the infinite product can easily be shown to converge using Eq. (1b)
below. It follows that
∞
n! = n ne − n 2 Pn ∏ (1 +
m=n
)
1 m +1 / 2
m
/e
(2a)
and we can show that the infinite product in this equation is negligible for large n. In particular,
each term in this product has m large and
(1 + m1 )m +1 / 2 / e = exp[(m + 12 ) ln(1 + m1 ) − 1]
= exp[( m + 12 )( 1 − 1 2 + 1 3 − L) − 1]
2m
3m
m
[
= exp
1
12 m 2
−
1
12 m 3
(1b)
]
+L
where use was made of Boas Eq. (1.13.4) in the second step, which applies since it is always true
that m1 ≤ 1. But for sufficiently large m, the argument of the exponential is negligibly small and it
follows that
lim (1 +
)
1 m +1 / 2
m
m →∞
=e
which actually converges quite rapidly (e.g., try it on a calculator for m = 1, 10, 100) and is in
and of itself a very useful result. Thus, Eq. (2a) becomes
n!~ n ne − n 2 Pn
(2b)
even for merely moderately large n.
The last and hardest step is to prove that P = π , so that Eq. (2b) gives the desired result. For
this purpose, we first derive Wallis’ formula for π starting from
π /2
∫ sin
lim
n →∞
0
π /2
∫ sin
2n
π /2
∫ sin
θ dθ
2 n +1
= lim
θ dθ
n →∞
π /2
∫ sin
2n
θ dθ
0
2n
θ cos( − θ ) dθ
=1
π
2
0
0
since for large n, sin θ is nonzero on (0,π /2) only for θ near π /2, in which case cos( π2 − θ ) ≈ 1.
2n
However,
π /2
Γ (n + 12 )Γ ( 12 )
1 B( n + 1 , 1 )
Γ (n + 1)
0
2
2 2 =
=
π /2
1 B( n + 1, 1 )
Γ (n + 1)Γ ( 12 )
2
2
2 n +1
∫ sin θ dθ
Γ (n + 23 )
∫ sin
2n
θ dθ
0
2
2
 Γ (n + 12 ) 

1 × 3 × 5 × L × (2 n − 1)
(n + 12 ) =  n
=
π  (n + 12 )


 2 1× 2 × 3 ×L× n
 Γ (n + 1) 
π 1 × 3 × 5 × L × (2 n − 1) 3 × 5 × 7 × L × (2 n + 1)
=
2 2 × 4 × 6 × L × 2n
2 × 4 × 6 × L × 2n
using the intermediate result of Boas Problem 11.5.1. Hence, in the limit as n → ∞ we have
π 2⋅2 4⋅4 6⋅6
=
×
×
×L
2 1⋅ 3 3 ⋅ 5 5 ⋅ 7
which is Wallis’ formula. Finally, consider the following quantity
2
2
2
 (2 n)! 
 1 × 2 × 3 × L × 2n 
1 × 3 × 5 × L × (2 n − 1) 
2n  n 2  = 2n 
2
n
=
2
 2 × 4 × 6 × L × 2 n 
 (2 n!) 
 (2 × 4 × 6 × L × 2 n) 
 2 ⋅ 2 × 4 ⋅ 4 × 6 ⋅ 6 × L × (2 n − 2 ) ⋅ (2 n − 2 ) × 2 n 
=
1 ⋅ 3 × 3 ⋅ 5 × 5 ⋅ 7 × L × (2 n − 3) ⋅ (2 n − 1) × (2 n − 1) 
−1
so that
2
2
 (2 n)! 
lim 2 n  n 2  =
n →∞
π
 (2 n!) 
using Wallis’ formula. (Compare Boas Problem 11.11.4.) On the other hand, Eq. (2b) implies
2
2
(2 n)4 n e −4 n 4 Pn
 (2 n)! 
2 n  n 2  ~ 2 n 4 n 4 n −4 n 2 2 =
P
2 n e 4P n
 (2 n!) 
for large n, and by comparing the last two equations we see that we must have P = π as required.
See Mermin for a discussion of how we can go on to refine Stirling’s approximation by
substituting Eq. (1b) into Eq. (2a).