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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.

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Is it possible to prove by induction that the theorem is true? – Paulo Argolo Nov 30 '10 at 18:35

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 $\frac{1}{12n+1}<r_{n}<\frac{1}{12n}$

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

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

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