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: 
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)
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.