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.
 A: Applying the Taylor development to the function $f(A)=\frac{1}{\sqrt{1-A^2}}$
$$
\begin{align}
f(A) &= (1-A^2)^{-\frac{1}{2}} \\
&= 1 + \frac{1}{1!} \left(-\frac{1}{2}\right)(-A^2)+ \frac{1}{2!}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)(-A^2)^2 +...+\frac{1}{n!}\left(\prod_{i=0}^{n-1}\left(-\frac{1}{2}-i\right)\right)(-A^2)^n+...\\
&=1 + \sum_{n=1}^{+\infty}\left(\frac{1}{n!}(-A^2)^n\prod_{i=0}^{n-1}\left(-\frac{1}{2}-i\right)\right) \\
&=1 + \sum_{n=1}^{+\infty}\frac{(2n-1)!!}{2^nn!}(A^2)^n \\
&=1 + \sum_{n=1}^{+\infty}\left(\frac{(2n-1)!!}{2^nn!}\times\frac{(2n)!!}{2^nn!}\right)(A^2)^n \\
&=1 + \sum_{n=1}^{+\infty}\frac{(2n)!}{(2^n)^2(n!)^2}(A^2)^n \\
&=\sum_{n=0}^{+\infty}\frac{(2n)!}{(n!)^2}\left(\frac{A^2}{2}\right)^n \\
\end{align}
$$
Q.E.D
A: Let
$$y=\sum_{n=1}^\infty \frac{(2n)!}{(n!)^2} x^{2n}=\sum_{n=1}^\infty\binom{2n}{n}x^{2n}.$$
Before we begin, note that
$$\binom{2n+2}{n+1}=\frac{2(2n+1)}{n+1}\binom{2n}{n}.$$
The strategy I will use here is to manipulate the sum in two different ways to make the $n$th coefficient equal to $\binom{2n+2}{n+1}$ instead of $\binom{2n}{n}$. Here goes:
(Note: In order to make this solution as clear as possible I have included every  step that is relevant; the solution is actually quite straightforward and if you persevere you will reasonably quickly get to the end :-) )
$$\begin{align}y&=\sum_{n=1}^\infty \frac{(2n)!}{(n!)^2} x^{2n}\\
\implies xy&=\sum_{n=1}^\infty \frac{(2n)!}{(n!)^2} x^{2n+1}\\
\implies \frac{d}{dx}(xy)=y+x\frac{dy}{dx}&=\sum_{n=1}^\infty (2n+1)\binom{2n}{n}x^{2n}\\
\implies xy+x^2\frac{dy}{dx}&=\sum_{n=1}^\infty (2n+1)\binom{2n}{n}x^{2n+1}\\
\implies\int xy+x^2\frac{dy}{dx}~dx&=C_1+\sum_{n=1}^\infty\frac{(2n+1)}{2(n+1)}\binom{2n}{n}x^{2n+2}=C_1+\frac{1}{4}\sum_{n=1}^{\infty}\binom{2n+2}{n+1}x^{2n+2} .\end{align}$$
However, we can also write
$$\sum_{n=1}^\infty \binom{2n+2}{n+1}x^{2n+2}=y-2x^2.$$
Hence,
$$\begin{align}\int xy+x^2\frac{dy}{dx}~dx&=C_1+\frac{1}{4}(y-2x^2)\\
\implies xy+x^2\frac{dy}{dx}&=\frac{1}{4}\frac{dy}{dx}-x\\
\implies \frac{dy}{dx}=(y+1)\cdot\frac{x}{\frac{1}{4}-x^2}.\end{align}$$
Separating the variables:
$$\begin{align}\int\frac{1}{1+y}~dy&=\int\frac{x}{\frac{1}{4}-x^2}dx\\
\implies \ln(1+y)&=C_2-\frac{1}{2}\ln\left(\frac{1}{4}-x^2\right)\\
\implies\ln(1+y)&=\ln\frac{1}{2}+\ln\frac{1}{\sqrt{\frac{1}{4}-x^2}}\\
\implies 1+y&=e^{\ln\frac{1}{2}}\cdot e^{\ln\frac{1}{\sqrt{\frac{1}{4}-x^2}}}\\
&=\frac{1}{\sqrt{1-4x^2}}\\
\implies \binom{2\cdot0}{0}x^{2\cdot0}+\sum_{n=1}^\infty\binom{2n}{n}x^{2n}&=\frac{1}{\sqrt{1-4x^2}}\\
\implies \sum_{n=0}^\infty\binom{2n}{n}x^{2n}&=\frac{1}{\sqrt{1-4x^2}}.\end{align}$$
Plug in $x=\displaystyle\frac{\lvert A\rvert}{2}$ and you have your answer.

I hope that helps. If you have any questions please don't hesitate to ask :)
A: 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*}
