Talk:Stirling's Formula/Refinement

The old version of the statement was seemingly a mistake

The old statement was:

A refinement of Stirling's Formula is:

$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {12 n} }$

I assume this was a mistake because this is not a refinement at all.

This is trivial because:

$1 + \dfrac 1 {12 n} \sim 1$

We can say anything like:

$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {13 n} }$

$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {14 n} }$

$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {n^2} }$

$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + e^{-n} }$

--Usagiop (talk) 23:22, 19 October 2022 (UTC)

Go and tell Don Knuth. --prime mover (talk) 05:06, 20 October 2022 (UTC)

Surprisingly, he is using the symbol $\approx$ without defining what it means. Not nice. --Usagiop (talk) 11:07, 20 October 2022 (UTC)