Stirling Formula

Harry Potter

Can you prove this Formula ? Smile

: $\lim_{n \to +\infty}{\frac{n^{n}.e^{-n}.\sqrt{2\pi.n}}{n!}} = 1$

**Posted:** Fri Nov 06, 2009 12:38 am

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kenn4000

http://www.sosmath.com/calculus/sequence/stirling/stirling.html

**Posted:** Fri Nov 06, 2009 1:25 am

npizzo · P versus NP Offline Joined: 22 Aug 2006 Posts: 26

Proofs of Stirling's formula are all over the place. Here's a loose proof for $z \in \mathbb{N}$ but it's not too hard to extend this argument for all $z \in \mathbb{C}$
Consider
$\Gamma(z + 1) = \int_0^{\infty} e^{ - t}t^{z}dt = \int_0^{\infty} e^{ - t + z\ln(t)}dt \quad \text{for}\ z \gg 1$
Let
$f(z) = - t + z\ln(t) \implies f(t_o) = 0 = - 1 + \frac {z}{t_o} \implies t_o = z$
Now we let's center our origin around this maxima. Making the substitution $t = z(1 + u) \implies$
$\Gamma(z + 1) = z\int_{ - 1}^{\infty} e^{ - z + zu + + z \ln z + zu - zu^2/2 + ...}du$
where we have used the Taylor Expansion for $\ln(1 + u)$ .
Note that $u$ is on the order of $O(\frac {1}{\sqrt {z}})$ so we can say the lower limit can be approximated as $- \infty$ and for a first order expansion we have.
$\implies \Gamma(z + 1) \sim z\left(\frac {z}{e}\right)^z \int_{ - \infty}^{\infty} e^{ - zu^2/2}du$

$\therefore \boxed{\Gamma(z + 1) = z! \sim \left(\frac {z}{e}\right)^z \sqrt {2 \pi z }}$

**Posted:** Fri Nov 06, 2009 10:37 am

Kent Merryfield

Here's some crude calculus estimates:

Approximately integrating $\int_1^n\ln x\,dx$ by the trapezoid rule (and using the fact that the function is concave downwards) we get

$\int_1^n\ln x\,dx>\sum_{k=2}^{n-1}\ln k+\frac12\ln n=\sum_{k=1}^{n}\ln k-\frac12\ln 2$

which rearranges to

$n!<e\sqrt{n}n^ne^{-n}.$

The next effort is to approximately integrate $\int_{\frac12}^{n+\frac12}\ln x\,dx$ by midpoint Riemann sums. I can rearrange this to get

$n!>\sqrt{2}\left(1+\frac1{2n}\right)^{n+\frac12}\sqrt{n}n^ne^{-n}.$

And that's approximately $\sqrt{2e}\sqrt{n}n^ne^{-n}.$

These methods aren't strong enough to get the $\sqrt{2\pi},$ but they're getting into the general neighborhood.

**Posted:** Fri Nov 06, 2009 10:56 am