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.
 A: HINT:
$$\frac{n^2}{n!}=\frac{n}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$$
A: If$$\lim_{n\to\infty}\log f(n)=-\infty$$
Then
$$\lim_{n\to\infty} f(n)=0$$
We can note that
$\log n!=\log 1+\log 2+\log 3+\cdots+\log n$
$\log n^2=2\log n$
A: This is another, rather sneaky way of doing it. 
$ Theorem :$ For $L$ belongs to set $R$ If $lim |(x_{n+1}/x_{n})| = L$ then sequence, 
$x_n$ is convergent if $L<1$, and
$x_n$ is divergent if $L>1$
However it must be noted that if $L=1$, then we don't have any information about its convergence 
A: You already got one hint, with exact equalities, here's one with inequalities:
You can safely assume that $n>4$, in which case
$$n!=n\cdot (n-1)\cdot (n-2)\cdot (n-3)!$$
and for $n>4$, you have that $n-3>1$, so $$n!\geq n(n-1)(n-2)$$
meaning that $$\frac{n^2}{n!} < \frac{n^2}{n(n-1)(n-2)}=\frac{n^2}{n^3-3n^2+2n}=\frac{1}{n-3+\frac{2}{n}}.$$
Can you continue from here?
A: For $n>2$,
$$
\frac{n^2}{n!}=
\frac{n}{n}\frac{n}{n-1}\frac{1}{n-2}\frac{1}{(n-3)!}<
2\frac{1}{n-2}
$$
because $\dfrac{n}{n-1}<2$ and $\dfrac{1}{(n-3)!}\le 1$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\color{#f00}{\lim_{n \to \infty}{n^{2} \over n!}} & =
\lim_{n \to \infty}{n \over \pars{n - 1}!} =
\lim_{n \to \infty}
{\pars{n + 1} - n \over \bracks{\pars{n + 1} - 1}! - \pars{n - 1}!} =
\lim_{n \to \infty}{1 \over \pars{n - 1}\pars{n - 1}!} = \color{#f00}{0}
\end{align}
A: 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
