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Discussing Stirling's approximation, Wolfram Mathworld article mentions a modification of it due to Gosper: instead of the usual

$$n!\approx n^ne^{-n}\sqrt{2n\pi},$$

we have a tiny addition of $\frac13$ to $2n$ under the radical sign:

$$n!\approx n^ne^{-n}\sqrt{\left(2n+\frac13\right)\pi}.$$

This radically improves the precision of approximation for all $n\ge0$. But all the places I've seen discussing it don't explain how it was obtained. Wolfram Mathworld just says about Gosper's modification

...a better approximation to n! (i.e., one which approximates the terms in Stirling's series instead of truncating them)...

but it doesn't make me understand how this change was derived.

So, how to derive this modification? And is there a further improvement to the whole Stirling series, or is it just "whole series stuffed into one expression"?

Ruslan
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    see the section on speed of convergence & error estimates ... https://en.wikipedia.org/wiki/Stirling%27s_approximation ... your formula is probably the same as including the $1/(12n)$ term. – Donald Splutterwit Mar 27 '17 at 11:16
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    Gosper's garden of brilliant ideas has many startling things like this :-) – ShreevatsaR Mar 28 '17 at 07:00
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    there is also this formula by Ramanujan. https://www.johndcook.com/blog/2012/09/25/ramanujans-factorial-approximation/ – user25406 Mar 28 '17 at 17:54

3 Answers3

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As it is said it the Wolfram article, Gosper's formula approximates the Stirling series instead of truncating it.

To see that, let's take a look at the $\sqrt{2n+\frac{1}{3}}$ term which is itself a series:

$$ \sqrt{2n+\frac{1}{3}} = \sqrt{2n} \cdot\sqrt{1+\frac{1}{6n}} = \sqrt{2n} \cdot \left( 1 + \frac{1}{12n} + O \left( \frac{1}{n^2}\right) \right) $$

So in the in the end you have:

$$ \frac{n!}{n^n e^{-n}\sqrt{2 n \pi}} = 1 + O \left( \frac{1}{n}\right) $$

While:

$$ \begin{align}\frac{n!}{n^n e^{-n}\sqrt{ \left( 2 n + \frac{1}{3} \right) \pi}} & = \frac{n!}{n^n e^{-n}\sqrt{2 n \pi}} \cdot\frac{1}{\sqrt{1+\frac{1}{6n}}} \\ & = \left( 1 + \frac{1}{12n} + O \left( \frac{1}{n^2}\right) \right) \cdot \left( 1 - \frac{1}{12n} + O \left( \frac{1}{n^2}\right) \right) \\ & = 1 + O \left( \frac{1}{n^2}\right) \end{align} $$

That explains why this approximation is much better than the simple truncation.

Ruslan
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yultan
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Stirling series is

$$n!\sim n^ne^{-n}\sqrt{2n\pi}\left(1+\frac1{12n}+\cdots\right).$$

Taking only terms up to and including $\frac1{12n}$, we have:

$$\sqrt{2n}\left(1+\frac1{12n}\right)=\sqrt{2\left(n+\frac16+\frac1{12^2n}\right)}=\sqrt{2n+\frac13+\frac1{72n}}.$$

Neglecting the $\frac1{72n}$ term in the radical (since it vanishes as $n\to\infty$), we recover the result for the radical by Gosper:

$$\sqrt{2n+\frac13}.$$

Ruslan
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1

An elementary proof of Gosper's approximation may be found here https://javeeh.net/StirlingsFormula

JAVeeh
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    Please put at least a brief summary of your proof here. Answers that need links to make sense are not useful on Stack Exchange sites. Please see [Your answer is in another castle: when is an answer not an answer?](https://meta.stackexchange.com/questions/225370/your-answer-is-in-another-castle-when-is-an-answer-not-an-answer) – PM 2Ring Mar 28 '17 at 13:05