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:
$$ \lim_{n\to\infty} \frac{n^{\ln \ln \ln n}}{\lceil(\ln n)\rceil!} $$
On
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 $0$.
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} $