Find the smallest value of $x$ such that the sequence converges to $0$.

Question

Define the real sequence:

$$a_n=\frac{n^n}{(n!)^x}$$

for $n\in\mathbb{N}$ (not including zero).

Find the smallest positive value of $x$ such that $\lim_{n\to \infty}a_n=0$.

Does such a positive real number $x$ exist? If not, is $1$ the infimum of the set of all $x$ for which the limit equals $0$?

Answer 1

Let use Stirling's approximation

$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

then

$$\frac{n^n}{(n!)^x}\sim n^n\left(\frac{e^n}{n^n\sqrt{2 \pi n}}\right)^x$$