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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?

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    $\begingroup$ Have you tried Stirling's Approximation. In particular, $\ln n!=n\ln n-n+O(\log n)$? $\endgroup$ – Michael Burr Apr 28 '16 at 14:59
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    $\begingroup$ Also, are you dealing with $\ln(n^{\ln n})$ or $(\ln n)^{\ln n}$? $\endgroup$ – Michael Burr Apr 28 '16 at 15:01
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    $\begingroup$ @MichaelBurr Stirling is overkill here (although, if one can take it for granted and build on it, why not) -- you only need something much looser, namely $\ln(n!) > c n\ln n$ for some positive $c>0$. And this can be achieved with very simple arguments (see e.g. my answer). $\endgroup$ – Clement C. Apr 28 '16 at 15:34
  • $\begingroup$ Taking the log of $(\ln n)^{\ln n}/n!$ gives $\ln n\ln\ln n-\ln(n!)$, not $(\ln n)^2-\ln(n!)$. $\endgroup$ – Barry Cipra Apr 28 '16 at 16:01
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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.

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$$a_{e^n}=\frac{n^n}{\Gamma(1+e^n)}=\frac{n}{e^n}\frac{n}{e^n-1}...\frac{n}{e^n-n+1}\frac{1}{\Gamma(1+e^n-n)}\le\left(\frac{n}{e^n-n+1}\right)^n\frac{1}{\Gamma(1+e^n-n)}$$ Each factor converges to $0$.

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  • $\begingroup$ $a_{e^n}$ is not continuous. $\endgroup$ – Kenny Lau Apr 28 '16 at 15:25
  • $\begingroup$ Well it is, but the RHS is not. $\endgroup$ – Kenny Lau Apr 28 '16 at 15:25
  • $\begingroup$ It's the same if we use $\lim\limits_{n\to\infty}$ or $\lim\limits_{e^n\to\infty}$. $\endgroup$ – user90369 Apr 28 '16 at 15:30
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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).

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    $\begingroup$ ........... $n$! $\endgroup$ – Kenny Lau Apr 28 '16 at 15:02
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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, $\frac{\ln 2}{2}n < \frac{1}{6} n \ln n$.

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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 $a=1/2,b=e$?

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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!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola May 26 '16 at 16:06
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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$

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