continued fraction $F(x)$ that is a generating function of central binomial coefficients Given the following continued fraction
$$F(x) =\cfrac{1}{x+\cfrac{2^2(2^2-1)}{6x+\cfrac{3^2(3^2-1)}{12x+\cfrac{4^2(4^2-1)}{20x+\cfrac{5^2(5^2-1)}{30x+\ddots}}}}}=\frac{1}{\sqrt{x^2+4}}$$
Then $$\frac{1}{x}F\left(\frac{1}{x}\right)=\sum^{\infty}_{n=0}(-1)^{n} \binom{2n}{n} x^{2n}$$ 
Where $\binom{2n}{n}$ are central binomial coefficients

How do we prove that the given continued fraction is a generating function of central binomial coefficients?

 A: $F(x)$ can be rewritten as $\displaystyle\;\frac{1}{\frac{2}{P(x)} - x}$ where
$\displaystyle\;\def\CF{\mathop{\LARGE\mathrm K}}
P(x) = \cfrac{1\cdot 2}{1\cdot 2 x + \cfrac{ (1 \cdot 2)(2\cdot 3)}{2\cdot 3 x + \cfrac{(2\cdot 3)(3\cdot 4)}{3\cdot 4 x + \ddots} }} 
$.
The CF $P(x)$ has the form $
\displaystyle\;
\CF_{\ell=1}^{\infty} \frac{\alpha_\ell\gamma_{\ell-1}}{\beta_\ell}
$
where
$\gamma_0 = 1$ and 
$\displaystyle\;
\begin{cases}
\alpha_\ell &= \ell (\ell+1)\\
\beta_\ell  &= \ell (\ell+1) x\\
\gamma_\ell &= \ell (\ell+1)
\end{cases}
$ for $\ell > 0$.
In general, CF of this form is invariant when we scale all $\alpha_\ell, \beta_\ell, \gamma_\ell$ by same factor for all $\ell > 0$.
If we scale $\alpha_\ell, \beta_\ell, \gamma_\ell$ by $\ell(\ell+1)$ for all $\ell > 0$, we find
$$P(x) = \CF_{\ell=1}^{\infty} \frac{1}{x} = \cfrac{1}{x + \cfrac{1}{x + \cfrac{1}{x + \ddots}}}$$
The CF at RHS is well known. It is not hard to verify its convergents have the form:
$$
\CF_{\ell=1}^{n} \frac{1}{x} = \lambda_{+}\lambda_{-}\frac{\lambda_{-}^n - \lambda_{+}^n}{\lambda_{-}^{n+1} - \lambda_{+}^{n+1}}
\quad\text{ where }\quad
\lambda_{\pm} = \frac{-x \pm \sqrt{x^2+4}}{2}$$
When $x > 0$, we have $|\lambda_{-}| > |\lambda_{+}|$.
This implies
$$\begin{align}
P(x) &= \lim_{n\to\infty} \CF_{\ell=1}^{n} \frac{1}{x} = \lambda_{+} = \frac{\sqrt{x^2+4} - x}{2}\\
\implies 
F(x) &= \frac{1}{\frac{4}{\sqrt{x^2+4}-x} - x}
= \frac{1}{(\sqrt{x^2+4}+x) - x} = \frac{1}{\sqrt{x^2+4}}
\end{align}
$$
Recall for any $\alpha \in \mathbb{C}$ and $|z| < 1$, we have
$$\frac{1}{(1-z)^{\alpha}} = \sum_{k=0}^\infty \frac{(\alpha)_k}{k!} z^k$$
where $(\alpha)_k = \alpha(\alpha+1)\cdots(\alpha+k-1)$. When $\alpha = \frac12$, the coefficient for $z^k$ becomes
$$\frac{(\frac12)_k}{k!} = \frac{1}{k!}\prod_{j=0}^{k-1}\left(j + \frac12\right)
= \frac{(2k-1)!!}{2^k k!} = \frac{(2k)!}{4^k(k!)^2} = \frac{1}{4^k} \binom{2k}{k}$$
From this, we find
$$\frac{1}{\sqrt{1-4z}} = \sum_{k=0}^\infty \binom{2k}{k} z^k
\quad\implies\quad 
\frac1x F\left(\frac1x\right) 
= \frac{1}{\sqrt{1+4x^2}} 
= \sum_{k=0}^\infty (-1)^k \binom{2k}{k} x^{2k}$$
