# Stirling's Formula

An important formula in applied mathematics as well as in probability is the Stirling's formula known as

where is used to indicate that the ratio of the two sides goes to 1 as n goes to . In other words, we have

or

## Proof of the Stirling's Formula

First take the log of n! to get

Since the log function is increasing on the interval , we get

for . Add the above inequalities, with , we get

Though the first integral is improper, it is easy to show that in fact it is convergent. Using the antiderivative of (being ), we get

Next, set

We have

Easy algebraic manipulation gives

Using the Taylor expansion

for -1 < t < 1, we get

This implies

We recognize a geometric series. Therefore we have

From this we get
1.
the sequence is decreasing;
2.
the sequence is increasing.
This will imply that converges to a number C with

and that C > d1 - 1/12 = 1 - 1/12 = 11/12. Taking the exponential of dn, we get

The final step in the proof if to show that . This will be done via Wallis formula (and Wallis integrals). Indeed, recall the limit

Rewriting this formula, we get

Playing with the numbers, we get

Using the above formula

we get

Easy algebra gives

since we are dealing with constants, we get in fact . This completes the proof of the Stirling's formula.

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