Steven R. Dunbar
Department of Mathematics
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University of Nebraska-Lincoln
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Topics in
Probability Theory and Stochastic Processes
Steven R. Dunbar
Stirling’s Formula Derived from Elementary Sequences
and Series
Rating
Mathematicians Only: prolonged scenes of intense rigor.
1
Section Starter Question
What is the geometric summation formula? How can you use the geometric
sum formula to derive the series expansion for log(1 + x) ? What do you need
to know about the geometric sum formula in order to justify its use to derive
the series expansion for log(1 + x) ?
Key Concepts
1. Stirling’s Formula, also called Stirling’s Approximation, is the asymptotic relation
√
n! ∼ 2πnn+1/2 e−n .
2. The formula is useful in estimating large factorial values, but its main
mathematical value is in limits involving factorials.
Vocabulary
1. Stirling’s Formula, also called Stirling’s Approximation, is the asymptotic relation
√
n! ∼ 2πnn+1/2 e−n .
2
Mathematical Ideas
Stirling’s Formula, also called Stirling’s Approximation, is the asymptotic
relation
√
n! ∼ 2πnn+1/2 e−n .
The formula is useful in estimating large factorial values, but its main mathematical value is in limits involving factorials. Another attractive form of
Stirling’s Formula is:
n n
√
n! ∼ 2πn
.
e
An improved inequality version of Stirling’s Formula is
√
√
2πnn+1/2 e−n+1/(12n+1) < n! < 2πnn+1/2 e−n+1/(12n) .
(1)
See Stirling’s Formula in MathWorld.com.
Here we rigorously derive Stirling’s Formula using elementary sequences
and series expansions of the logarithm function, based on the sketch in
Kazarinoff [2] which is in turn based on the note by Nanjundiah [3].
A weak form of Stirling’s Formula
Theorem 1.
√
n
n
e
Remark. This form of Stirling’s Formula is weaker than the usual form since
it does not give direct estimates on n!. √On the other hand,it avoids the
determination of the asymptotic constant 2π which usually requires Wallis’s
Formula or equivalent. For many purposes of estimation or limit taking this
version of Stirling’s Formula is sufficient, and the proof is elementary. The
proof is taken from [5, pages314-315].
n! ∼
Proof. Start from the series expansion for the exponential function and then
estimate:
n
nn−1
nn
n
n2
n
e = 1 + + ... +
+
1+
+
+ ...
1!
(n − 1)!
n!
n + 1 (n + 1)(n + 2)
j
∞ n
nn nn X
<n +
n!
n! j=0 n + 1
nn
= (2n + 1)
n!
3
On the other hand, en > nn /n! by dropping all but the nn /n! term from the
series expansion for the exponential. Rearranging these two inequalities
nn
(2n + 1)nn
< n! <
.
en
en
Now take the nth root of each term, and use the fact that
n → ∞.
√
n
2n + 1 → 1 as
Stirling’s Formula from sequences and series
n!
. Then
Let an = nn+1/2
e−n
1/2) log(1 + 1/n) − 1 .
an
an+1
n+1 n+1/2 −1
e
n
=
and log
an
an+1
= (n +
Lemma 2. For |x| < 1,
log
1+x
1−x
=2
∞
X
k=0
1
.
(2k + 1)x2k+1
Proof. Left as an exercise.
Note then that
1
log 1 +
= log
n
1+
1−
1
2n+1
1
2n+1
!
=2
∞
X
k=0
1
.
(2k + 1)(2n + 1)2k+1
Then
log
an
an+1
=
=
2n + 1
2
∞
X
k=1
2
∞
X
k=0
1
(2k + 1)(2n + 1)2k+1
!
∞
−1
X
1
1
=
.
2k+1
(2k + 1)(2n + 1)
(2k + 3)(2n + 1)2k+2
k=0
Now coarsely estimating the denominators
∞
X
an
1
1
log
≤
.
2
an+1
2(2n + 1) k=0 (2n + 1)2k
4
Lemma 3.
log(an+1 ) < log (an ) <
1
1
−
+ log(an+1 )
12n 12(n + 1)
Proof. Let f (x) = (x + 1/2) log(1 + 1/x) − 1, so log
an
an+1
= f (n). Observe
2x+1
that f (x) → 0 as x → ∞ and f 0 (x) = log(1 + 1/x) − 2x
Because
2 +2 .
0
f (x) < 0 for x > 1 (proof left as an exercise) f (x) is decreasing
from
an
(3/2) log(2) − 1 ≈ 0.0397 to 0 as x increases. Hence 0 < log an+1 .
By Lemma 2
∞
X
an
1
1
log
=
.
2
an+1
3(2n + 1) k=0 (2n + 1)2k
The sum is a geometric sum, so
an
1
1
1
0 < log
<
=
2
2
an+1
3(2n + 1) 1 − (1/(2n + 1) )
12n(n + 1)
Expand 1/(12n(n + 1)) in partial fractions and add log(an+1 ) throughout
to obtain
log(an+1 ) < log(an ) <
1
1
−
+ log(an+1 )
12n 12(n + 1)
1
Define xn = log(an ) − 12n
, so Lemma 3 shows that xn is an increasing
sequence, xn < xn+1 . That is,
xn = log(an ) −
1
1
< log(an+1 ) −
= xn+1 .
12n
12(n + 1)
(2)
Define yn = log(an ), and then the left-side inequality in Lemma 3 shows that
yn is a decreasing sequence that is, yn+1 < yn . By the definition of xn and
yn , xn < yn and |xn − yn | = 1/(12n), so |xn − yn | → 0 as n → ∞. Therefore
sup xn = inf yn and call the common value λ. By continuity, limn→∞ an = eλ .
Using elementary properties of limits
(limn→∞ an )2
(an )2
e = lim an =
= lim
.
n→∞
n→∞ a2n
limn→∞ a2n
λ
5
(3)
However,
(an )2
=
a2n
s 2 · 4 . . . (2n − 2) · 2n
2
.
n 1 · 3 . . . (2n − 3) · (2n − 1)
(4)
The easy demonstration is left as an exercise.
Lemma 4 (Wallis’ Formula).
(2n) · (2n) . . . 2 · 2
π
lim
= .
n→∞
(2n + 1) · (2n − 1) · (2n − 1) . . . 3 · 3 · 1
2
Proof. See the proofs in Wallis Formula.
Using the continuity of the square root function
!
r
r
1
(2n) · (2n − 2) . . . 4 · 2
π
.
lim
=
n→∞
2n + 1
(2n − 1) · (2n − 3) . . . 5 · 3 · 1
2
Now multiplying both sides by 2 and rewriting the leading square root
sequence, obtain
!
r
√
2 2n
(2n) · (2n − 2) . . . 4 · 2
= 2π .
lim
n→∞
n 2n + 1
(2n − 1) · (2n − 3) . . . 5 · 3 · 1
Then since
r
lim
n→∞
2n
=1
2n + 1
equation 4 is
s √
2 · 4 . . . (2n − 2) · 2n
2
eλ = lim an =
= 2π .
n→∞
n 1 · 3 . . . (2n − 3) · (2n − 1)
Equivalently, unwrapping the definition of an =
Stirling’s Formula
√
n! ∼ 2πnn+1/2 e−n .
n!
nn+1/2 e−n
this is exactly
1
Using the definitions xn = log(an ) − 12n
and yn = log(an ), the inequality xn < yn , and the least upper bound and greatest lower bound limit in
equation (3) we can express Stirling’s Formula in inequality form
√
√
2πnn+1/2 e−n < n! < 2πnn+1/2 e−n+1/(12n) .
This is almost as good as the inequality (1).
6
Figure 1: Example of the lemma with α = 1 (green) and α = 0 (red).
Remark. This proof of Stirling’s Formula and the inequality (1) is the easiest,
the shortest and the most elementary of the proofs that I present. These are
all definite advantages. The main disadvantage of this proof is that it requires
the form of Stirling’s Formula before starting, in order to create the sequence
an which is the main object of the proof.
Stirling’s Formula from Wallis’ Formula
Lemma 5. If α ∈ R, then the sequence
n+α
1
aα (n) = 1 +
n
is decreasing if α ∈ [ 12 , ∞), and increasing for n ≥ N (α) if α ∈ (−∞, 21 ).
Remark. This lemma is based on a remark due to I. Schur, see [4], problem
168, on page 38 with solution on page 215
7
x+α
(defined on [1, ∞))
Proof. The derivative of the function f (x) = 1 + x1
is
x+α 1
1
x+α
0
f (x) = 1 +
ln 1 +
−
.
x
x
x(x + 1)
Let
x+α
1
−
g(x) = ln 1 +
x
x(x + 1)
then
g 0 (x) =
(2α − 1)x + α
x2 (x + 1)2
and limx→∞ g(x) = 0. It follows that g(x) < 0 and so f 0 (x) < 0 when
α
). The
α ≥ 1/2 and x ≥ 1, and f 0 (x) > 0 when α < 1/2 and x ≥ max(1, 1−2α
monotonicity of aα (n) follows.
From the lemma, for every α ∈ (0, 1/2) there is a positive integer N (α)
such that
k+α
k+1/2
1
1
<e< 1+
1+
k
k
for all k ≥ N (α). As a consequence, we obtain
2n−1
Y
k=n
1
1+
k
k+α
n
<e <
2n−1
Y
k=n
1
1+
k
k+1/2
.
Rearrange the products with telescoping cancelations, using the upper
8
bound on the right as an example.
k+1/2
2n−1
Y
1
1+
k
k=n
n+1/2 n+1+1/2
2n−1+1/2
1
1
1
= 1+
1+
··· 1 +
n
n+1
2n − 1
n+1/2 n+1+1/2
2n−1+1/2
n+2
2n
n+1
···
=
n
n+1
2n − 1
1/2 1/2
1/2 n n+1
2n−1
n+1
n+2
2n
n+1
n+2
2n
=
···
·
···
n
n+1
2n − 1
n
n+1
2n − 1
1/2 2n
(2n)2n−1
=
·
n
nn · (n + 1) · · · (2n − 1)
(2n)2n−1
= 21/2 · n
n · (n + 1) · · · (2n − 1)
22n−1 · nn−1
.
= 21/2 ·
(n + 1) · · · (2n − 1)
Multiply by the last fraction by n!/nn and Write more compactly,
n! 1/2
22n−1 · nn−1
·2 ·
.
nn
(n + 1) · · · (2n − 1)
2n−1 · 2n · n!
= 21/2 ·
n · (n + 1) · · · (2n − 1)
(2n)!!
= 21/2 ·
(2n − 1)!!
This uses the “double factorial” notation n!! = n · (n − 2) · · · 4 · 2 if n is even,
and n!! = n · (n − 2) · · · 3 · 1 if n is odd. The lower bound product on the left
is similar.
That is, after multiplying through by n!/nn+1/2 ,
2α
(2n)!!
n!en
21/2
(2n)!!
√ ·
< n+1/2 < √ ·
n
n (2n − 1)!!
n (2n − 1)!!
for all n ≥ N (α). Using the Wallis Formula,
√
√
n!en
n!en
2α π ≤ lim inf n+1/2 ≤ lim sup n+1/2 ≤ 2π.
n→∞ n
n→∞ n
9
Stirling’s formula follows by passing to the limit as α → 1/2.
Remark. This proof can be extended to an asymptotic formula for the Gamma
function using log-convexity of the Gamma function. See [1].
Sources
The weak form of Stirling’s Formula is taken from [5]. The first sequence
proof is mostly adapted from the sketch in Kazarinoff [2] which is in turn
based on the note Nanjundiah [3]. The second sequence proof using derivatives and monotonicity is adapted from [1].
Problems to Work for Understanding
1. Show that
√
n
2n + 1 → 1 as n → ∞.
2. Show that for |x| < 1,
∞
X
1
1+x
.
log
=2
2k+1
1−x
(2k
+
1)x
k=0
3. Let f (x) = (x + 1/2) log(1 + 1/x) − 1. Show that f (x) decreases to 0
as x → ∞.
4. Show that
s 2
2 · 4 . . . (2n − 2) · 2n
(an )
=
.
a2n
n 1 · 3 . . . (2n − 3) · (2n − 1)
2
x+α
5. Show that the derivative of the function f (x) = 1 + x1
(defined on
[1, ∞) ) is
x+α 1
1
x+α
0
f (x) = 1 +
ln 1 +
−
.
x
x
x(x + 1)
10
Reading Suggestion:
References
[1] Dorin Ervin Dutkay, Constantin P. Niculescu, and Florin Popvici. Stirling’s formula and its extension for the Gamma function. American Mathematical Monthly, 120:737–740, October 2013. available as
http:dx.doi.org/10.4169/amer.math.monthly.120.08.737.
[2] Nicholaus D. Kazarinoff. Analytic Inequalities. Holt, Rinehart and Winston, 1961.
[3] T. S. Nanjundiah. Note on Stirling’s formula. American Mathematical
Monthly, 66:701–703, 1959.
[4] G. Pólya and G. Szegö. Problems and Theorems in Analysis I: Series,
Integrals, Theory of Functions. Classics in Mathematics. Springer Verlag,
1998. Repring of the 1978 version.
[5] S. Saks and A. Zygmund. Analytic Functions. Elsevier Publishing, 1971.
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Email to Steve Dunbar, sdunbar1 at unl dot edu
Last modified: Processed from LATEX source on January 6, 2014
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