# 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?

• Have you tried Stirling's Approximation. In particular, $\ln n!=n\ln n-n+O(\log n)$? – Michael Burr Apr 28 '16 at 14:59
• Also, are you dealing with $\ln(n^{\ln n})$ or $(\ln n)^{\ln n}$? – Michael Burr Apr 28 '16 at 15:01
• @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). – Clement C. Apr 28 '16 at 15:34
• Taking the log of $(\ln n)^{\ln n}/n!$ gives $\ln n\ln\ln n-\ln(n!)$, not $(\ln n)^2-\ln(n!)$. – Barry Cipra Apr 28 '16 at 16:01

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.

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

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

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 $Michael Burr suggested in the replies (Stirling). • ...........$n$! – Kenny Lau Apr 28 '16 at 15:02 It is easy to show that$\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,$\frac{\ln 2}{2}n < \frac{1}{6} n \ln n$. 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$a=1/2,b=e? Here is a "brute force" approach. Note that we can write \begin{align} \frac{(\log(n))^{\log(n)}}{n!}&=e^{\log(n)\log(\log(n))-\sum_{k=1}^n\log(k)}\\\\ &=e^{\log(n)\log(\log(n))-n\log(n)-\sum_{k=1}^n\log(k/n)}\\\\ &\le e^{\log(n)\log(\log(n))-n\log(n)+n} \end{align} where the inequality is due to the fact that\frac1n \sum_{k=0}^n\log(k/n)\le \int_0^1 \log(x)\,dx=-1$. 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! • Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola May 26 '16 at 16:06 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\$