Computing the series $\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \Big(\frac{|A|}{2}\Big)^{2n}$ I want to know how to compute the closed form for the series
$$\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \Big(\frac{|A|}{2}\Big)^{2n}, \quad |A| < 1,$$
with or without special functions. When I plugged this into Mathematica, it gave me
$$\frac{1}{\sqrt{1 - |A|^2}}.$$
I also would really, really like to know if there are exercises/lessons all in one place (it's a bit hard to search the SE for something like this) or texts out there that can help me learn methods of doing these computations. I know I won't be able to always find closed forms of expressions like this, but I'm always amazed at what I see folks are able to compute here on Stack Exchange.
I thought to use a modified Bessel function $I_\alpha(z)$ with $\alpha = 0$,
$$I_0(z) = \sum_{k=0}^\infty \frac{1}{(k!)^2} \left(\frac{z}{2}\right)^{2k}$$
and perhaps differentiate and then evaluate at $|A|$, but I haven't figured out how to manipulate that. Any suggestions, general and specific?
Note: For some context, I am generally trying to compute norms for squeezed coherent states (quantum mechanics application) which have the form $\psi(z) = e^{Az^2/2} e^{-|z|^2/2}$, $z \in \mathbb{C}$, and I wind up with series like this all the time.

Just for reference, I guess, I finally did go back and take the suggestion of @samario28 and use the generalized binomial theorem,
$$ (1 + z)^\alpha = \sum_{n=0}^\infty \frac{(\alpha)_n}{n!} z^n = \sum_{n=0}^\infty \binom{\alpha}{n} z^n,$$
where $(\alpha)_n$ denotes the falling factorial. Just for good measure, a ratio test of $a_n = (2n)!/(n!)^2 z^{2n}$ will show that radius of convergence is for $|z| < 1/2$, which corresponds to our values for $|A|$. We have
\begin{align*} \binom{2n}{n} &= \frac{(2n)!}{n!n!} \\
&= \frac{2n (2n-1)(2n-2) \cdots 3 \cdot 2 \cdot 1}{n!n!} \\
&= \frac{2^{2n} n (n-\tfrac{1}{2}) (n-1) (n- \frac{3}{2}) \cdots \frac{3}{2} \cdot 1 \cdot \frac{1}{2} \cdot 1}{n!n!} \\
&= \frac{2^{2n} (n-\tfrac{1}{2}) (n- \frac{3}{2}) \cdots \frac{3}{2} \cdot \frac{1}{2}}{n!} \\
&= \frac{2^{2n} (-1)^n (\tfrac{1}{2}-n) (\frac{3}{2}-n) \cdots (-\frac{3}{2}) \cdot (-\frac{1}{2})}{n!} \\
&= \frac{2^{2n} (-1)^n (-\frac{1}{2})(-\frac{1}{2}-1) \cdots (-\frac{1}{2}-n+2)(-\frac{1}{2} - n+1)}{n!} \\
&= \frac{2^{2n} (-1)^n (-\frac{1}{2})_{n}}{n!}.
\end{align*}
Thus, $\alpha = -1/2$, and so
\begin{align*} \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \left(\frac{|A|}{2}\right)^{2n} &= \sum_{n=0}^\infty \frac{\big(-\!\frac{1}{2}\big)_n}{n!} \cdot 2^{2n} (-1)^n \left(\frac{|A|}{2}\right)^{2n} \\
&= \sum_{n=0}^\infty \frac{\big(-\!\frac{1}{2}\big)_n}{n!} \cdot (-|A|^2)^n \\
&= \frac{1}{\sqrt{1 - |A|^2}}.
\end{align*}
