Question: is this right ?
$$
\sum\_{n=1}^{\infty} \frac{n^2} {n!}
$$
i need to use quotient criterion 

$$
\lim\_{n\to\infty} \frac{\frac{n^2}{n!}}{ \frac{1}{n!}} = \lim\_{n\to\infty} {\frac{n^2}{n!}} { \frac{n!}{1}} = \lim\_{n\to\infty} {n^2} = \infty
$$

so $$ \lim\_{n\to\infty} \frac{\frac{n^2}{n!}}{ \frac{1}{n!}} $$ equals $\infty$ and $$\sum\_{n=1}^{\infty} \frac{1} {n!} $$ is divergent it means that  

$$ \sum\_{n=1}^{\infty} \frac{n^2} {n!}$$ is divergent

Answer: 
Can you show that $\dfrac{n^2}{n!}<\dfrac{1}{n^2}$ for sufficiently large $n$? 

Then use the monotonic property of the sum from that $n$ onwards. The other $n$'s less than that particular $n$ contribute only to a finite sum.

EDIT: Since the question used the words "Limit Comparison", consider comparing with $\dfrac{1}{(n-2)!}$.

Notice that $$\lim\_{n\to\infty}\dfrac{\dfrac{n^2}{n!}}{\dfrac{1}{(n-2)!}}=\lim\_{n\to\infty}\dfrac{n}{n-1}=1$$

Note that $\dfrac{1}{n!}<\dfrac{1}{n^2}$ for all $n>3$. Thus $\sum\_{n=3}^\infty\dfrac{1}{(n-2)!}$ converges. Hence $\sum\_{n=3}^\infty \dfrac{n^2}{n!}$ converges, and hence $\sum\_{n=1}^\infty\dfrac{n^2}{n!}$ converges.
This doesn't work!

In the Limit Comparison Test, you need
$$
\lim\_{n\to \infty} \frac{a\_n}{b\_n}
$$
to be a positive real number (and not infinity) when comparing $\sum a\_n$ and $\sum b\_n$. 

Consider the example where
$$
a\_n = \frac{1}{n} \quad\text{and}\quad b\_n = \frac{1}{n^2}.
$$
Here $\sum a\_n$ is divergent and $\sum b\_n$ is convergent, but
$$
\lim\_{n\to \infty} \frac{a\_n}{b\_n} = \infty.
$$

I don't know how you would use the Limit Comparison Test here. I see now that Landon does know. So you should see his answer!

As a sidenote, note that
$$
\sum\_{n=1}^{\infty} \frac{1}{n!}
$$
is convergent and not divergent.