How to prove $\lim\limits_{n \to \infty} \frac{n}{\log_2 n!} = 0$ I believe this cannot be handled by just analytical transformations and L'Hôpital's rule (there is no derivate of n!...)
How can I prove that:
$\lim\limits_{n \to \infty} \frac{n}{\log_2 n!} = 0$
My only attempt so far at breaking this one was:
$\lim\limits_{n \to \infty} \frac{n}{\log_2 n!} = \lim\limits_{n \to \infty} \frac{\log_2 2^n}{\log_2 n!} = \lim\limits_{n \to \infty} \log_{n!} 2^n$
It is easy to prove that
$\lim\limits_{n \to \infty} \frac{2^n}{n!} = 0$
Because:
$\lim\limits_{n \to \infty} \frac{2^n}{n!} = \lim\limits_{n \to \infty} \binom{2}{1}\binom{2}{2}\binom{2}{3}...\binom{2}{n} < \lim\limits_{n \to \infty} \frac{4}{n} = 0$
Because both $2^n > 0$ and $n! > 0$
But I don't know how to use it as proof for limit
 A: Using Cesaro-Stolz:
$$\lim_{n \to \infty} \frac{n}{\log_2 n!} =\lim_{n \to \infty} \frac{n+1-n}{\log_2 (n+1)!-\log_2 n!}=\lim_{n \to \infty} \frac{1}{\log_2 (n+1)} = 0$$
A: $n(n-1)(n-2).....1 \ge (n/2)^{n/2}$;
$\dfrac{n}{\log_2 n!}\le \dfrac{n}{(n/2)(\log_2 n-\log_2 2)}=$
$\dfrac{2}{\log_2 n -1}$.
Take the limit.
A: Note that for every $k >0$ we get $2^{kn} = \left(2^k \right)^n < n!$ for almost all $n$. In particular, $\log_2(n!) > kn$ for almost all $n$. This implies the claim.
A: Doing some basic simplifications with our limit, we can see that:
$$\lim_{n\to\infty}\frac{n}{\log_2{(n!)}}\\= \lim_{n\to\infty}\frac{n}{\left(\frac{\ln(n!)}{\ln(2)}\right)}\\= \ln(2)\cdot\lim_{n\to\infty}\frac{n}{\ln{(n!)}}$$ In order to prove that this limit approaches $0$, we need to prove that $\ln{(n!)}>n$ for sufficiently large $n$. This inequality can be simplified to:
$$n!>e^{n}$$
Note that the rate of which $n!$ grows as $n$ grows, whereas the growth of $e^{n}$ is constant. From this, we can show that $n!>e^n$, for sufficiently large $n$. We can now apply this inequality to where we left off in our limit simplification, hence proving that:
$$\lim_{n\to\infty}\frac{n}{\log_2{(n!)}}=0$$
