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.
Another straightforward use of the limit test can be done in small pieces:

* Show that $n^2\lt 4\cdot(n-1)^2$ for sufficiently large $n$ (hint: this is equivalent to $n\lt 2(n-1)$ for positive $n$; you should be able to solve that inequality algebraically)
* This then gives $\sum\_{n=n\_0}^\infty \frac{n^2}{n!}\lt 4\sum\_{n=n\_0}^\infty \frac{(n-1)^2}{n!}$.
* Now, $\frac{(n-1)^2}{n!}=\frac{(n-1)^2}{n\cdot(n-1)\cdot(n-2)!}\lt\frac{(n-1)^2}{(n-1)\cdot(n-1)\cdot(n-2)!}=?$

Can you fill in the rest of the details from here?

(You should also be able to transform this into a more direct use of the limit comparison test proper — the key concept is to shift the series to get rid of that quadratic term up top)