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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!

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  • $\begingroup$ You may find it easier to take the exponential function of the expressions above, find the limit, and then take the log of the limit you find. You can do this because the exponential and logarithm functions are continuous (ie, they commute with limits). $\endgroup$ Oct 29, 2012 at 16:20
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    $\begingroup$ 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 $n=\log(e^{n})$ $\endgroup$ Oct 29, 2012 at 16:27

7 Answers 7

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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:

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

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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 $0$.

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    $\begingroup$ What is $\psi$? $\endgroup$ Oct 29, 2012 at 17:15
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    $\begingroup$ Is it really imaginable that someone who has trouble with those limits can even have heard of the digamma function? $\endgroup$
    – commenter
    Oct 29, 2012 at 18:12
  • $\begingroup$ @commenter:He said"any tool can be used"! $\endgroup$ Oct 29, 2012 at 20:31
  • $\begingroup$ @commenter It is not. $\endgroup$
    – Did
    Oct 30, 2012 at 10:18
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    $\begingroup$ @commenter: Nothing is wrong with this approach. However, I already gave him an alternative approach. $\endgroup$ Oct 30, 2012 at 11:35
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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 $$ \lim_{n\to\infty}\frac{\log(n!)}{\log(n)^{\log(n)}}=\lim_{n\to\infty}\frac{n\log(n)}{\log(n)^{\log(n)}}\stackrel{n\to e^n}{=}\lim_{n\to\infty}\frac{e^nn}{n^n}=\lim_{n\to\infty}\left(\frac{2e}{n}\right)^n\frac{n}{2^n}=0 $$

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

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    $\begingroup$ Stoltz looks like L'Hopital: $\lim\limits_{n\to\infty}\frac{a_n}{b_n}=\lim\limits_{n\to\infty}\frac{\Delta a_n}{\Delta b_n}$ $\endgroup$
    – robjohn
    Jan 5, 2013 at 4:36
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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 $0$.

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

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    $\begingroup$ how does this apply to either question? $\endgroup$
    – robjohn
    Jan 5, 2013 at 1:20
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For the first one, use that $\log(n!) \sim n \log n$ as $n \to \infty$ to conclude that $$ \lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{\log(n!)} = \lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{n\log n} = \lim_{n\to\infty}\frac{3}{\sqrt{n}\log n} = 0 $$

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    $\begingroup$ This answers half the question. +1/2 rounded up :-) $\endgroup$
    – robjohn
    Jan 5, 2013 at 4:30

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