Calculate the limit $\lim_{n \to \infty}\frac{ \ln(n)^{(\ln n)}}{n!}$ I wonder what the limit $\lim_{n \to \infty}\frac{ \ln n^{\ln n}}{n!}$ would be equal to. It is well known that the factorial function grow faster than an exponential but slower than $n^n$. But how about a combination of $\ln $ (natural logarithm) and exponential? I guess the answer is $0$ since for $e$ the value is quite small. If I show that the logarithm of the expressions tends to $-\infty$ then I would be done.  Using laws of logarithm I can write $(\ln n)^2-\ln(n)!=(\ln n)^2(1-\frac{\ln(n!)}{(\ln n)^2})$. Now I need to know the limit of $\frac{\ln(n!)}{(\ln n)^2}$. Any suggestions?  
 A: It's easy to see that $\ln n\lt n/2$ and $n!\gt(n/2)^{n/2}$ for $n\gt1$.  Thus
$${(\ln n)^{\ln n}\over n!}\lt{(n/2)^{\ln n}\over(n/2)^{n/2}}={1\over(n/2)^{(n/2)-\ln n}}$$
and the latter tends to $0$ for any number of reasons.
A: $$a_{e^n}=\frac{n^n}{\Gamma(1+e^n)}=\frac{n}{e^n}\frac{n}{e^n-1}...\frac{n}{e^n-n+1}\frac{1}{\Gamma(1+e^n-n)}\le\left(\frac{n}{e^n-n+1}\right)^n\frac{1}{\Gamma(1+e^n-n)}$$
Each factor converges to $0$.
A: Let
$$a_n=\frac{\left(\ln n\right)^{\ln n}}{n}$$
Then
$$\ln a_n=\ln\frac{\left(\ln n\right)^{\ln n}}{n}=\ln\left(\left(\ln n\right)^{\ln n}\right)-\ln n=\ln n\cdot\ln\ln n-\ln n=\ln n\cdot\left(\ln\ln n-1\right)$$
It is easy to see that $\ln a_n\to\infty$, so also $a_n\to\infty$
EDIT:
I just noticed that 
$$a_n=\frac{\left(\ln n\right)^{\ln n}}{n!}$$
I would look again at $\ln a_n$ and use the trick $Michael Burr suggested in the replies (Stirling).
A: It is easy to show that $\ln (n!) > \frac{1}{3} n \ln n$.$^{(\dagger)}$ Using this, rewrite
$$
0 < \frac{(\ln n)^{\ln n}}{n!} < \frac{(\ln n)^{\ln n}}{e^{ \frac{1}{3} n \ln n}} = e^{(\ln n)\ln \ln n -  \frac{1}{3} n \ln n} 
= e^{(\ln n)( \ln \ln n -  \frac{1}{3} n )} \xrightarrow[n\to \infty]{} 0 
$$
where the limit follows from observing that the exponent goes to $\infty\cdot -\infty = -\infty$, and continuity of the exponential.

Proof of $(\dagger)$: 
$$
\ln(n!) = \sum_{j=1}^n \ln j > \sum_{j=n/2+1}^n \ln j > \frac{n}{2} \ln \frac{n}{2} = \frac{1}{2} n \ln n - \frac{\ln 2}{2}n 
$$
and for $n$ sufficiently big, $\frac{\ln 2}{2}n < \frac{1}{6} n \ln n$.
A: You are dealing with 
$$\frac{\exp((\ln n)^2)}{n!}$$
when $n$ is large. (If I'm not misinterpreting it, you mean $(\ln n)^{\ln n}$ rather than $\ln(n^{\ln n})$.)
For an elementary approach, two facts, which every sensible calculus textbook should include, might be helpful here:
1). For any $a>0$ (however terribly small), and any $\epsilon>0$, there exists $N\in \Bbb Z^+$ depending on $a$ and $\epsilon$ such that $n\ge N$ implies $\ln n \le \epsilon n^a$.
2). For any $b>1$ (however terribly large), and any $\epsilon>0$, there exists $N\in \Bbb Z^+$ depending on $b$ and $\epsilon$ such that $n\ge N$ implies $b^n\le \epsilon n!$.
Hint: what happens when you let $a=1/2,b=e$?
A: Here is a "brute force" approach.  Note that we can write
$$\begin{align}
\frac{(\log(n))^{\log(n)}}{n!}&=e^{\log(n)\log(\log(n))-\sum_{k=1}^n\log(k)}\\\\
&=e^{\log(n)\log(\log(n))-n\log(n)-\sum_{k=1}^n\log(k/n)}\\\\
&\le e^{\log(n)\log(\log(n))-n\log(n)+n}
\end{align}$$
where the inequality is due to the fact that $\frac1n \sum_{k=0}^n\log(k/n)\le \int_0^1 \log(x)\,dx=-1$.
Now, for any $\alpha>0$ the logarithm function is bounded by
$$\log(n)\le \frac{n^\alpha -1}{\alpha}$$
Then choosing, say $\alpha =1/2$, it is easy to see that
$$\lim_{n\to \infty}e^{\log(n)\log(\log(n))-n\log(n)+n}=0$$
And we are done!
A: If , $\ln(n)^{\ln n}  =  \left ( \ln n \right )^2 $ then it is easy to see that it will be smaller than $n !$ for bigger $n$, because $n! = n \cdot (n - 1) \cdot (n- 2)\cdots$ Whereas $\ln n \lt n$ for every $n > 0$. You just need first three term of $n!$ to cancel out $ n^2$
