Does it converge and find possibly the limit of: $\frac{n^n}{n!}$

There is probably a much more "elegant solution than mine:

$$\lim\limits_{n\to\infty}\frac{n^n}{n!}=\frac{1 \cdot 4 \cdot 27 \cdot 256... }{1 \cdot 2 \cdot 6 \cdot 24...} \le n \cdot \frac{n}{2} \cdot \frac{n}{3}... $$ and it converges to infinity.

I would really appreciate any kind of help.

Solution 1:

We can apply D'Alembert criterion with $a_n=\frac{n^n}{n!}$:

$\lim\limits_n\frac{a_{n+1}}{a_n}=\lim\limits_n \frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\lim\limits_n \left(\frac{n+1}{n}\right)^n=\lim\limits_n\left(1+\frac{1}{n}\right)^n=e$.

Because this limit is $>1$, then $a_n\to\infty$.

Solution 2:

The easiest way is just to note that $n!=\prod\limits_{k=2}^n k\leq n^{n-1}$. Thus:

$\frac{n^n}{n!}\geq n$

And so the limit is infinity.