1
$\begingroup$

I need to calculate:

$\lim\limits_{n \to \infty} \frac{n\log_2n}{\log_2 n!}$

Wolfram says it is 1. However I got different result:

First I will use Cesaro-Stolz to remove $n!$

$\lim\limits_{n \to \infty} \frac{n\log_2n}{\log_2 n!} = \lim\limits_{n \to \infty} \frac{(n+1)\log_2(n+1) - n\log_2n}{\log_2 (n+1)! - \log_2 n!} = \lim\limits_{n \to \infty} \frac{(n+1)\log_2(n+1) - n\log_2n}{\log_2 (n+1)}$

Now I can apply L'Hôpital's rule:

$\lim\limits_{n \to \infty} \frac{(n+1)\log_2(n+1) - n\log_2n}{\log_2 (n+1)}=\lim\limits_{n \to \infty} \frac{\log2(n+1)+\frac{1}{\ln2}-\log2n+\frac{1}{n\ln2}}{\frac{1}{(n+1)\ln2}}$

And I apply it again:

$\lim\limits_{n \to \infty} \frac{\log2(n+1)+\frac{1}{\ln2}-\log2n+\frac{1}{n\ln2}}{\frac{1}{(n+1)\ln2}} = \lim\limits_{n \to \infty} \frac{\frac{1}{(n+1)\ln2}-\frac{1}{n\ln2}-\frac{1}{n^2\ln2}}{\frac{1}{(n+1)^2\ln2}} = \lim\limits_{n \to \infty} \frac{(n+1)^2\ln2}{(n+1)\ln2}-\frac{(n+1)^2\ln2}{n\ln2}-\frac{(n+1)^2\ln2}{n^2\ln2} = \lim\limits_{n \to \infty} (n+1) - \frac{(n+1)^2}{n} - \frac{(n+1)^2}{n^2} = \lim\limits_{n \to \infty} \frac{n^3+n^2-n^3-2n^2-n-n^2-2n-1}{n^2} = \lim\limits_{n \to \infty} \frac{-2n^2-3n-1}{n^2} = -2$

I am probably missing something obvious, but having triple checked my calculation I don't see any obvious mistakes...

If there is an easier way to calculate that limit, I would gladly accept

$\endgroup$
3
$\begingroup$

Using Cesaro-Stolz:

$$\lim_{n\to \infty} \frac{n\log_2 n}{\log_2 n!} = \lim_{n\to \infty} \frac{(n+1)\log_2 (n+1)-n\log_2 n}{\log_2(n+1)}= 1+\lim_{n\to \infty}\frac{n\log_2(n+1)-n\log_2n}{\log_2(n+1)}$$$$=1+\lim_{n\to \infty} \frac{\log_2 (n+1)^n-\log_2n^n}{\log_2(n+1)}=1+\lim_{n\to \infty} \frac{\log_2 \left(1+\dfrac{1}{n}\right)^n}{\log_2(n+1)}=1+\frac{\log_2 e}{\infty}=1$$

$\endgroup$
1
$\begingroup$

I would start with $n!=1\cdot2\cdot3\cdot...\cdot n$, then $\log_2(n!)$ can be expressed as:

$\log_2(n!)=\log_2(1\cdot2\cdot3\cdot...\cdot n)=\log_2(1)+\log_2(2)+\log_2(3)+...+\log_2(n) \le n\log_2(n)$.

Hence, when $n\rightarrow\infty$ we will have $\log_2(n!) \approx n\log_2(n)$, so the limit will result with $1$.

$\endgroup$
0
$\begingroup$

$$ \frac{n\log_2(n)}{\log_2 (n!)}= \frac{n\log(n)}{\log( n!)}$$ Now using Stirling approximation $$\log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({n}\right)\right)+\frac{1}{12n}+O\left(\frac{1}{n}\right)$$ Then $$\frac{\log( n!)}{n\log(n)}=1-\frac{1}{\log (n)}+\frac{\log (2 \pi n)}{2 n \log (n)} \to 1$$ $$\frac{n\log(n)}{\log( n!)}=\frac 1 {\frac{\log( n!)}{n\log(n)} }\to 1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.