Calculus of One and Several Variables

9.1 Stirling’s Formula

Stirling’s formula is a very useful approximation for n!. I am going to give a presentation of this which follows that in the calculus book by Courant. A very different treatment is in [32].

Let

∫ n n An ≡ 1 ln(x)dx = x ln (x )− x|1 = nlnn − n+ 1.

Form a uniform partition of

, each sub-interval having length 1, and sum the areas of the trapezoids which result. The k^th trapezoid which lies under the graph of y = lnx has area equal to

ln-(k+-1)+-ln(k) 2

Here is a picture of one of these.

Thus, summing the areas of the inscribed trapezoids and denoting this approximate area by T_n, it equals

n−1 T = ∑ ln(k+-1)+-ln(k)= 1(ln (n!)+ ln((n− 1)!)) = ln(n!)− 1lnn n k=1 2 2 2

Here this comes from the observation that ∑ _k=1ⁿ⁻¹ ln

= ln from the laws of logarithms. Of course, it follows from the picture that the approximation is smaller than A_n. Let a_n = A_n − T_n.

Now consider the above picture. Approximate the area between the graph of ln

and the inscribed trapezoid with the difference in area between the two trapezoids in the above picture. Note that the area of the larger trapezoid is ln. (why?) Thus

n∑−1( ( 1) ln(k+ 1)+ ln(k)) 0 ≤ an ≤ ln k+ 2 − -------2------- k=1

Consider the term of the series. It equals  

Thus Since a_n is increasing as n increases, this implies that lim_n→∞a_n = β exists. Now this implies that e^a_n also converges to some number α. Hence

A nlnn−n+1 n −n√ -- lim e-n-= lim e---------= lim n-e----ne= α n→∞ eTn n→ ∞ eln(n!)− 12lnn n→ ∞ n!

Letting c = eα⁻¹, it follows from the above that

n −n√ -- lim n-e----nc = 1. n→ ∞ n!

This has proved the following lemma.

There are various ways to show that this constant c equals

. Using integration by parts, it follows that whenever n is a positive integer larger than 1,

∫ π∕2 n − 1∫ π∕2 sinn(x)dx = ----- sinn−2 (x)dx 0 n 0

Proof: Consider the first formula in the case where m = 1. From beginning calculus,

∫ π∕2 2 π 1π sin (x)dx = 4-= 22- 0

so the formula holds in this case. Suppose it holds for m. Then from the above reduction identity and induction,

The second claim is proved similarly. ■

Now from the above identities,

∫π∕2 2m 2 --0--sin---(x)dx- = π(2m + 1)-(2m-−-1)-⋅⋅⋅32-⋅12- ∫π0∕2sin2m+1 (x)dx 2 (2m )2 (2m − 2)2⋅⋅⋅22

which implies

∫ π 1 (2m )2(2m − 2)2⋅⋅⋅22 π0∕2sin2m (x)dx 2-= 2m-+-1-(2m-−-1)2⋅⋅⋅32-⋅12--∫π∕2--2m+1------ 0 sin (x)dx

From the reduction identity,

∫π∕2sin2m (x) dx ∫π∕2sin2m (x)dx 1 ≤ ∫π0∕2------------= ----0∫-π∕2-------------≤ 2m-+-1 0 sin2m+1 (x)dx 22mm+1-0 sin2m−1(x)dx 2m

It follows

It follows that

2m π ∕2 2m-+-1 ≤ ----(2m−-2)2⋅⋅⋅22- ≤ 1 2m (2m −1)2⋅⋅⋅32⋅12

and so

(2m − 2)2⋅⋅⋅22 π lim 2m -------2---------= -- m →∞ (2m − 1) ⋅⋅⋅32 ⋅12 2

This is sometimes called Wallis’s formula. This implies

22m−-2((m-−-1)!)2-- π- mlim→∞ 2m (2m − 1)2⋅⋅⋅32 ⋅12 = 2

Now multiply on the bottom and top by

²²2² = 2^2m2 to obtain

It follows that

22m (m!)2 ∘ π- lim ----1∕2------= -- (9.1) m→ ∞ (2m ) (2m )! 2

(9.1)

Now with this result, it is possible to find c in Stirling’s formula. Recall

lim -----m!------= 1 m→ ∞ mm+ (1∕2)e− mc

It follows that

-----(2m-)!------ mli→m∞ (2m )2m+ (1∕2)e−2mc = 1

Therefore, from 9.1,

which shows that c = . This proves Stirling’s formula.