My result of $\lim\limits_{n \to \infty} \frac{n\log_2n}{\log_2 n!}$ is different than Wolfram, where had I made a mistake? 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
 A: Using Cesaro-Stolz:
$$\lim_{n\to \infty} \frac{n\log_2 n}{\log_2 n!} = \lim_{n\to \infty} \frac{(n+1)\log_2 (n+1)-n\log_2 n}{\log_2(n+1)}= 1+\lim_{n\to \infty}\frac{n\log_2(n+1)-n\log_2n}{\log_2(n+1)}$$$$=1+\lim_{n\to \infty} \frac{\log_2 (n+1)^n-\log_2n^n}{\log_2(n+1)}=1+\lim_{n\to \infty} \frac{\log_2 \left(1+\dfrac{1}{n}\right)^n}{\log_2(n+1)}=1+\frac{\log_2 e}{\infty}=1$$
A: 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 $1$.
A: $$ \frac{n\log_2(n)}{\log_2 (n!)}= \frac{n\log(n)}{\log( n!)}$$ 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
$$\frac{\log( n!)}{n\log(n)}=1-\frac{1}{\log (n)}+\frac{\log (2 \pi  n)}{2 n \log (n)} \to 1$$
$$\frac{n\log(n)}{\log( n!)}=\frac 1 {\frac{\log( n!)}{n\log(n)} }\to 1$$
