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I am trying to calculate the following limits, but I don't know how: $\lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{\log(n!)}$ And the second one is $\lim_{n\to\infty}\frac{\log(n!)}{\log(n)^{\log(n)}}$

I don't need to show a formal proof, and any tool can be used.

Thanks!

  • 3
    The first one: Write $\lim_{n\to\infty}\frac{3\sqrt{n}}{\log n!}=\lim_{n\to\infty}\frac{3\sqrt{n}}{n}\cdot\frac{n}{\log n!}$ and use the product rule, knowing that $n=\log(e^{n})$2012-10-29

7 Answers 7

6

You can easily show that $2^n \leq n! \leq n^n$ for $n \geq 4$. The first inequality is a very standard induction proof, and the second inequality is straight-forward (you're comparing $1 \times 2 \times \dots \times n$ with $n \times n \times \dots \times n$).

From there, since $f(n) = \log n$ is an increasing function, you have that

$n\log(2) \leq \log(n!) \leq n\log(n)$

This tells you basically everything you will need. For example, for the first one:

$ \lim_{n \to \infty} \frac{3 \sqrt{n}}{n\log n} \leq \lim_{n \to \infty}\frac{3 \sqrt{n}}{\log(n!)} \leq \lim_{n \to \infty} \frac{3 \sqrt{n}}{n \log(2)}. $

2

If you use the fact that $ (\ln(x!))'= \psi(x+1)$, where $\psi(x)$ is the digamma function, and l'hopital's rule, then the first limit can be evaluated directly

$ \lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{\log(n!)}=\lim_{n \to \infty} \frac{3}{2}\frac{1}{\sqrt{n}\psi(n+1)}=0 \,,$

where the fact that $ \lim_{n\to \infty} \psi(n+1)=\infty \,, $

has been used.

For the second limit, recalling the asymptotic of $\ln(n!)$ $\ln(n!)=\sum_{k=1}^{n}\ln(k)\sim \int_{1}^{n} \ln(x)dx \sim n\ln(n)-n+1, $ we have $ \frac{\log(n!)}{\log(n)^{\log(n)}}\sim \frac{n\ln(n)-n+1}{\log(n)^{\log(n)}} \,.$

Making the change of variables $m=\ln(n)$ yields

$ \frac{\log(n!)}{\log(n)^{\log(n)}}\sim \frac{n\ln(n)-n+1}{\log(n)^{\log(n)}}=\frac{me^m-e^m+1}{m^m} \rightarrow 0 $

as $m\to \infty,$ since $m^m>me^m \,\,\, \forall m>4. $

Note: we can use $\ln(n!)\sim n\ln(n)-n+1 $ to prove the first limit goes to $0$.

  • 1
    @commenter: Nothing is wrong with this approach. However, I already gave him an alternative approach.2012-10-30
2

Stirling's approximation yields $ \log(n!)=\left(n+\frac12\right)\left(\log(n)-1\vphantom{\frac12}\right)+\frac12\log(2\pi e)+O\left(\frac1n\right) $ which implies $ \lim_{n\to\infty}\frac{\log(n!)}{n\log(n)}=1 $ Then for the first limit $ \lim_{n\to\infty}\frac{3\sqrt n}{\log(n!)}=\lim_{n\to\infty}\frac3{\sqrt{n}\log(n)}=0 $ For the second limit $ \lim_{n\to\infty}\frac{\log(n!)}{\log(n)^{\log(n)}}=\lim_{n\to\infty}\frac{n\log(n)}{\log(n)^{\log(n)}}\stackrel{n\to e^n}{=}\lim_{n\to\infty}\frac{e^nn}{n^n}=\lim_{n\to\infty}\left(\frac{2e}{n}\right)^n\frac{n}{2^n}=0 $

1

For the first one you can solve it using stoltz lemma

$\lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{\log(n!)}= \lim _{n\rightarrow \infty} \frac{3\cdot \sqrt{n+1}-3\cdot \sqrt{n}}{\log((n+1)!)-\log n!}=\lim _ {n\rightarrow \infty }\frac{3 }{(\sqrt{n+1}+\sqrt{n})(\log(n+1))}=0$

For the second observe you can rewrite the denominator as

$\log(n)^{\log(n)}=e^{\log n\log (\log n)}=n^{\log (\log n)}$

$\lim_{n\to\infty}\frac{\log(n!)}{\log(n)^{\log(n)}}\leq \lim_{n\to\infty}\frac{n\log(n)}{\log(n)^{\log(n)}} =\lim_{n\to\infty}\frac{\log(n)}{n^{\log (\log n)-1}}$ Therefore $n>e^{e^3} \Rightarrow \log ( \log n)-1>2\Rightarrow n^{\log ( \log n)-1}>n^{2} \Rightarrow \frac{\log(n)}{n^{\log (\log n)-1}} \leq \frac{\log n}{n^2}$

So finally $\lim_{n\to\infty}\frac{\log(n!)}{\log(n)^{\log(n)}}\leq \lim _{ n \rightarrow \infty }\frac{\log n}{n^2}=0$

  • 1
    Stoltz looks like L'Hopital: $\lim\limits_{n\to\infty}\frac{a_n}{b_n}=\lim\limits_{n\to\infty}\frac{\Delta a_n}{\Delta b_n}$2013-01-05
1

Start with the fact that $\ln n!=\sum_{k=1}^n\ln k$. A look at the graph of $y=\ln x$ shows that

$\sum_{k=1}^n\ln k\ge\int_1^n\ln x~dx=\Big[x\ln x-x\Big]_1^n=n\ln n-n+1\;,$

so

$0\le\frac{3\sqrt n}{\ln n!}\le\frac{3\sqrt n}{n\ln n-n+1}<\frac3{(\ln n-1)\sqrt n}\;.$

Clearly $\displaystyle\lim_{n\to\infty}\frac3{(\ln n-1)\sqrt n}=0$, so $\displaystyle\lim_{n\to\infty}\frac{3\sqrt n}{\ln n!}=0$.

Similarly, $\ln n!=\sum_{k=1}^n\ln k\le\int_1^{n+1}\ln x~dx=(n+1)\ln(n+1)-n\;,$

so

$0\le\frac{\ln n!}{(\ln n)^{\ln n}}\le\frac{(n+1)\ln(n+1)-n}{(\ln n)^{\ln n}}\le\frac{(n+1)\ln(n+1)}{(\ln n)^{\ln n}}\le\frac{(n+1)^2}{(\ln n)^{\ln n}}\le\frac{4n^2}{(\ln n)^{\ln n}}$ for $n\ge 1$. And

$\frac{4n^2}{(\ln n)^{\ln n}}=\frac{4e^{2\ln n}}{(\ln n)^{\ln n}}=4\left(\frac{e^2}{\ln n}\right)^{\ln n}\to 0$

as $n\to\infty$, so the second limit is also $0$.

1

For the first one, use that $\log(n!) \sim n \log n$ as $n \to \infty$ to conclude that $ \lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{\log(n!)} = \lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{n\log n} = \lim_{n\to\infty}\frac{3}{\sqrt{n}\log n} = 0 $

  • 1
    This answers half the question. +1/2 rounded up :-)2013-01-05
1

Note that even $\ln{n!} \rightarrow \ln(n^n)$ $n!$ does not approche $n^n$

  • 3
    how does this apply to either question?2013-01-05