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