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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

PiotrK
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    You know immediately that your answer must be wrong because for $n$ large your sequence is positive, yet your limit is negative. – J. De Ro Jan 25 '20 at 23:03
  • Using these bounds leads very quickly to the limit $1$, and they are more elementary than Stirling: https://proofwiki.org/wiki/Lower_and_Upper_Bound_of_Factorial – Maximilian Janisch Jan 25 '20 at 23:04
  • using natural logarithm, $$ n \log n - n + 1 < \log n! < (n+1) \log (n+1) - n - 1 - 2 \log 2 + 2 $$ with the very elementary method at https://math.stackexchange.com/questions/3239790/the-asymptotic-behavior-of-n-ln-n-n/3239892#3239892 – Will Jagy Jan 26 '20 at 00:26

3 Answers3

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Using Cesaro-Stolz:

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

LHF
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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 $1$.

Anton Vrdoljak
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$$ \frac{n\log_2(n)}{\log_2 (n!)}= \frac{n\log(n)}{\log( n!)}$$ 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 $$\frac{\log( n!)}{n\log(n)}=1-\frac{1}{\log (n)}+\frac{\log (2 \pi n)}{2 n \log (n)} \to 1$$ $$\frac{n\log(n)}{\log( n!)}=\frac 1 {\frac{\log( n!)}{n\log(n)} }\to 1$$

Claude Leibovici
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