How to prove that n! have a higher order of growth than n^(log n)?

Question

I am aware that n^n have a higher order of growth than n!, but how about n^(log n)? Is there a way to get an alternative form of n^(log n) such that when taking the

lim n to infinity [alternative form(n^(log n))] / (n!)

it would equate to 0?

Answer 1

Hint: Since $\log(n)\le \sqrt{n}$ for all $n\ge N_0$ for some positive integer $N_0$, we get that:

$$\frac{n^{\log n}}{n!}=\frac{e^{(\log(n)^2)}}{n!}\le \frac{e^{n}}{n!}$$

Answer 2

Using a multiplicative variant of Gauss's trick we have: $$ (n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1) \ge n^n $$ So $$ \dfrac{n^{\log n}}{n!} \le \dfrac{n^{\log n}}{n^{n/2}} \le \dfrac{n^{n/4}}{n^{n/2}} = \dfrac{1}{n^{n/4}} \to 0 $$ because $\log n \le n/4$ for $n$ large enough ($n\ge 9$ actually).

Answer 3

Another hint: consider a series with the term $a_n=n^{\log n}/n!$ and use, e.g., D'Alambert's test to find out that your series converges. Hence the common term of the series must tend to zero.