Does the series $\sum^\infty_{n=1}\frac{n!}{\sqrt{(2n)!}}$ converge/diverge? 
Does the series $\displaystyle\sum^\infty_{n=1}\frac{n!}{\sqrt{(2n)!}}$ converge/diverge?

I used the ratio test but I'm not sure:
$\begin{align} \frac{\frac{(n+1)!}{\sqrt{(2n+2)!}}}{\frac{n!}{\sqrt{(2n)!}}}
&=\frac{n+1}{\sqrt{(2n+1)(2n+2)}}\\
&=\frac{n+1}{4(n+1)^2\sqrt{2n+1}}\\
&=\frac{1}{4\sqrt{2n+1}} \end{align}$
The limit of that is smaller than $1$ so the series does converge. Is it correct ?

This is in response to your question here.  It's too long for a comment so I'm posting it as an answer.
No, you can't take them out of the root like that.  One way to handle the denominator is by writing it like
$$
\begin{align}
\sqrt{(2n+1)\cdot(2n+2)} &= \sqrt{(2n+1)\cdot 2\cdot (n+1)} \\
&= \sqrt{2n+1} \sqrt{2} \sqrt{n+1},
\end{align}
$$
so that
$$
\frac{n+1}{\sqrt{(2n+1)(2n+2)}} = \frac{n+1}{\sqrt{2} \sqrt{2n+1} \sqrt{n+1}}.
$$
Now
$$
\frac{n+1}{\sqrt{n+1}} = \sqrt{n+1},
$$
so
$$
\frac{n+1}{\sqrt{2} \sqrt{2n+1} \sqrt{n+1}} = \frac{\sqrt{n+1}}{\sqrt{2}\sqrt{2n+1}}.
$$
To calculate the limit of this as $n \to \infty$, divide the numerator and the denominator by $\sqrt{n}$ to get
$$
\begin{align}
\frac{\frac{1}{\sqrt{n}}\sqrt{n+1}}{\frac{\sqrt{2}}{\sqrt{n}}\sqrt{2n+1}} &= \frac{\sqrt{1+\frac{1}{n}}}{\sqrt{2}\sqrt{2 + \frac{1}{n}}} \\
&\overset{n\to\infty}{\longrightarrow} \frac{\sqrt{1+0}}{\sqrt{2}\sqrt{2 + 0}} \\
&= \frac{1}{2}.
\end{align}
$$
