Using the Limit Comparison Test on $\sum_{n=1}^{\infty} \frac{n^2} {n!}$ 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
 A: 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.
A: 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)
A: I propose a simpler, direct approach:
\begin{align*}
\sum_{n=1}^{\infty}\frac{n^2}{n!}=&\,\sum_{n=1}^{\infty}\frac{n\times n}{n\times(n-1)!}=\sum_{n=1}^{\infty}\frac{n}{(n-1)!}=\sum_{n=1}^{\infty}\frac{n-1+1}{(n-1)!}=\sum_{n=1}^{\infty}\frac{n-1}{(n-1)!}+\sum_{n=1}^{\infty}\frac{1}{(n-1)!}\\
=&\,\sum_{n=0}^{\infty}\frac{n}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}=\underbrace{\frac{0}{0!}}_{=0}+\sum_{n=1}^{\infty}\frac{n}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}+\sum_{n=0}^{\infty}\frac{1}{n!}\\
=&\,\sum_{n=0}^{\infty}\frac{1}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}=2e.
\end{align*}
A: 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.
