There is the following limit, I would like to calculate:
$\lim_{n\rightarrow\infty}\frac{n!}{\left(n+1/6\right)!}$
I tried to use the Stirling approaximation formula
$n!\approx\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}$
After the substituion I have got a relatively complex formula. I suppose that it may be solved by the Hospital's rule...
Is it the right method for the limit computation, if we don't want to use the Gamma function?
Thanks for your help...
If you use Stirling's approximation you get $$\frac{n!}{\left(n+1/6\right)!}\approx\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}}{\approx\sqrt{2\pi (n+1/6)}\left(\frac{n+1/6}{e}\right)^{n+1/6}}=\frac{\sqrt{2\pi n}}{\sqrt{2\pi (n+1/6)}}\frac{1}{(\frac{n+1/6}{e})^{1/6}}(\frac{n}{n+1/6})^n$$. Now note that $$\frac{\sqrt{2\pi n}}{\sqrt{2\pi (n+1/6)}}\approx \frac{\sqrt{n}}{\sqrt{n+1/6}}\approx 1$$ and $$\frac{1}{(\frac{n+1/6}{e})^{1/6}}\approx 0$$ Now let us deal with the last bit of the product. For this we take the logarithm to get $$n(\ln(n)-\ln(n+1/6))=\frac{\ln(n)-\ln(n+1/6)}{1/n}$$ Hitting this with L'Hopital's, we get $$\lim_{n\to\infty}\frac{\ln(n)-\ln(n+1/6)}{1/n}=\lim_{n\to\infty}\frac{\frac{1}{n}-\frac{1}{n+1/6}}{-1/n^2}=\lim_{n\to\infty}-\frac{n^2/6}{n(n+1/6)}=-\frac{1}{6}.$$ Therefore, $\lim_{n\to\infty}(\frac{n}{n+1/6})^n=e^{-1/6}$. Putting all of this together, we get $$\lim_{n\to\infty}\frac{n!}{\left(n+1/6\right)!}=0$$
The derivation of Stirling's formula not relying on the gamma function depends on the fact that $n\in\mathbb{N}$. By using Stirling's formula for $n\notin\mathbb{N}$ we are analytically extending the formula to numbers for which the original derivation fails. As mentioned in the comments, secretly we are using the gamma function.
In any case, using Stirling's formula we find $$\begin{eqnarray*} \log\frac{n!}{(n+a)!} &\sim& \log \sqrt{2\pi n}(n/e)^n - \log \sqrt{2\pi(n+a)}((n+a)/e)^{n+a} \\ &=& \left(\log\sqrt{2\pi}+\frac{1}{2}\log n + n\log n - n\right) \\ && - \left(\log\sqrt{2\pi}+\frac{1}{2}\log(n+a) + (n+a)\log(n+a) - (n+a)\right) \\ &=& -\frac{1}{2}\log(1+a/n) - n\log(1+a/n) - a\log n - a\log(1+a/n) + a\\ &\sim& -a\log n, \end{eqnarray*}$$ where in the last step we use $\log(1+x)=x+O(x^2)$ for $x\ll 1$ and neglect terms of order $1/n$. Therefore, $n!/(n+a)! \sim 1/n^a\,(n\to\infty)$ and so, if $a>0$, $$\lim_{n\to\infty}\frac{n!}{(n+a)!} = 0.$$
