Calculating limit of factorials I am trying to show what a limit converges to. I have plotted it and can easily see it will converge to 0, you can also see it since the upper term is basically a constant. I am having a hard time rewriting the weird terms, so any tips or help to prove it will converge to 0 would be appreciated. See the limit here bellow:
$$ \lim_{n\to\infty} \frac{n^{\ln \ln \ln n}}{\lceil(\ln n)\rceil!} $$
 A: First, we will plug in the substitution $n=e^x$, clearly the limit for $n$ going to infinity is the same as for $x$ going to infinity. This gives
$$\lim_{n \rightarrow \infty}\frac{n^{\ln\ln\ln n}}{(\ln n)!} 
= \lim_{x \rightarrow \infty}\frac{e^{x\ln\ln x}}{x!}$$
Next we use the Sterling approximation formula for $x! \sim \sqrt{2\pi x}\left(\frac{x}{e}\right)^x$. Note that the ratio of these two expression goes to $1$ as $x$ goes to infinity. This gives 
$$ \lim_{x \rightarrow \infty}\frac{e^{x\ln\ln x}}{x!} 
= \lim_{x \rightarrow \infty}\frac{e^{x+x\ln\ln x}}{\sqrt{2\pi x} x^x}
= \lim_{x \rightarrow \infty}\frac{1}{\sqrt{2\pi x}}e^{x(1-\ln x + \ln\ln x)}$$
Now the initial fraction converges to $0$, the part $1-\ln x + \ln\ln x$ goes to minus infinity so the exponential also goes to zero. 
So you are correct, the limit is indeed $0$.
A: If
$a_n
=\frac{n^{\ln \ln \ln n}}{(\ln n)!}
=\frac{e^{\ln n \ln \ln \ln n}}{(\ln n)!}
$,
$b_n
=\ln(a_n)
=\ln n \ln \ln \ln n-\ln((\ln n)!)
$.
Since
$\ln(m!)
= m\ln m - m +O(\ln(m))
$,
$\ln((\ln n)!)
= \ln(n)\ln \ln(n) - \ln(n) +O(\ln(\ln(n)))
$,
so that
$\begin{array}\\
b_n
&=\ln n \ln \ln \ln n-\ln((\ln n)!)\\
&=\ln n \ln \ln \ln n-(\ln(n)\ln \ln(n) - \ln(n) +O(\ln(\ln(n))))\\
&=\ln n (\ln \ln \ln n-\ln \ln(n)) + \ln(n) +O(\ln(\ln(n))))\\
&\to -\infty\\
\text{so}\\
a_n
&\to 0\\
\end{array}
$
