Show that $\lim_{n \to \infty}\frac{n^2}{n!}=0$ Show that $\lim_{n \to \infty}\frac{n^2}{n!}=0$. Can you give me a hint? Thanks to all.
HINT:

$$\frac{n^2}{n!}=\frac{n}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$$
Not the most elegant but...

Hint: Use Stirling's Approxmation:

$$\lim \frac{n^2}{n!}$$

$$ = \lim \frac{n^2}{n!} \frac{n!}{\sqrt{2 \pi n} (\frac{n}{e})^n}$$

$$ = \lim \frac{n^2}{\sqrt{2 \pi n} (\frac{n}{e})^n}$$

Now compute:

$$\lim \frac{n^2}{\sqrt{n} (\frac{n}{e})^n}$$

$$ = \lim \frac{n^{3/2}}{(\frac{n}{e})^n}$$

Use LHR somewhere