Talk:Stirling's Formula

What is going on with the very last line of Proof 2? --Oliver (talk) 14:31, 7 May 2014 (UTC)


 * Timesing both sides by $\left({\sqrt n \left({\dfrac n e}\right)^n}\right)$ -- is this invalid in this context? --prime mover (talk) 16:53, 7 May 2014 (UTC)


 * Well firstly if you have $\displaystyle \lim_{x \to c} \frac {f(x)} {g(x)} = L$ and you want to multiply both sides by $g(x)$ you end up with $\displaystyle \lim_{x \to c} f(x) = \lim_{x \to c} L g(x)$, not just $\displaystyle \lim_{x \to c} f(x) = L g(x)$.


 * But in this case I wouldn't say that was the best form to leave it in because neither limit exists. It also doesn't emphasise what is important which is that they both tend to infinity at the same rate - I think it should be left in the form $\displaystyle \lim_{n \to \infty} \dfrac {n!} {\sqrt{2 \pi} n^n \sqrt n e^{-n} } = 1$ as this agrees with the definition of asymptotically equal. --Oliver (talk) 04:05, 8 May 2014 (UTC)


 * Are you in a position to make it rigorous? Many thx if you would. --prime mover (talk) 19:00, 8 May 2014 (UTC)

Stirling Formula Proof 2
Isn't it shorter to just estimate by Taylor formula (as done anyway) that $|d_{n+1}-d_n|$ is smaller than $c/n^2$ and, therefore, the sequence is convergent?