4
$\begingroup$

Is it even worth the theorem below?

For every positive integer $n$, there is a real number $r$, and $\frac{1}{12n+1} \lt r \lt\frac{1}{12n}$, such that: $$ n! = \sqrt{2n\pi}\left(\frac{n}{e}\right)^n e^r.$$

I saw this statement on some sites, but got no further details. I think the statement refers to an exact value of $n!$, not an approximation.

  • 0
    Is it possible to prove by induction that the theorem is true?2010-11-30

2 Answers 2

15

It is the strirling's aproximation See article

$n!\approx \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$

The error estimates are very interesting:

$n!=\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n} \cdot e^{r_{n}}$

With $\frac{1}{12n+1}

This is the strirling's aproximation with remainder

Edit: This aproximation is very useful, for example in the calculation of limits of sequences

For example,

$$\frac{\sqrt[n]{n!}}{n} \approx \frac{\sqrt[n]{\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}}}{n} = \frac{\left(\frac{n}{e}\right)\sqrt[n]{\sqrt{2\pi n}}}{n} = \frac{\sqrt[2n]{2\pi n}}{e} \longrightarrow \frac{1}{e}, ( n\longrightarrow \infty ) $$

Using the fact $\sqrt[n]{n} \longrightarrow 1, ( n\longrightarrow \infty )$

2

I think you are referring to the Stirling's approximation:

http://en.wikipedia.org/wiki/Stirlings_approximation