<s>[INST] A golden ratio series from a comic book The eighth installment of the Filipino comic series Kikomachine Komix features a peculiar series for the golden ratio in its cover:

That is,
$$\phi=\frac{13}{8}+\sum_{n=0}^\infty \frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}}$$
How might this be proven?
 [/INST] A few easy manipulations yield the equivalent series
$$\frac{13}{8}-\frac1{64}\sum_{n=0}^\infty \frac{(2n+1)!}{n!(n+2)!}\left(-\frac1{16}\right)^n$$
The problem, then, is to evaluate
$$f(z)=\sum_{n=0}^\infty \frac{(2n+1)!}{n!(n+2)!}z^n$$
Using the duplication formula for the factorial, we have
$$\begin{align*}
f(z)&=\sum_{n=0}^\infty 4^{n+\frac12}\frac{\left(n+\frac12\right)!}{\left(-\frac12\right)!}\frac{z^n}{(n+2)!}\\
&=2\sum_{n=0}^\infty \frac{\left(n+\frac12\right)!}{\left(-\frac12\right)!}\frac{(4z)^n}{(n+2)!}
\end{align*}$$
We can massage the series to a more familiar form like so:
$$\begin{align*}
f(z)&=\frac{2}{(4z)^2}\sum_{n=0}^\infty \frac{\left(n+\frac12\right)!}{\left(-\frac12\right)!}\frac{(4z)^{n+2}}{(n+2)!}\\
&=\frac{1}{8z^2}\sum_{n=2}^\infty \frac{\left(n-\frac32\right)!}{\left(-\frac12\right)!}\frac{(4z)^n}{n!}\\
&=-\frac{1}{4z^2}\sum_{n=2}^\infty \frac{\left(n-\frac32\right)!}{\left(-\frac32\right)!}\frac{(4z)^n}{n!}\\
&=-\frac{1}{4z^2}\sum_{n=2}^\infty \binom{n-\frac32}{n}(4z)^n\\
\end{align*}$$
We then flip the binomial coefficient to give
$$\begin{align*}
f(z)&=-\frac{1}{4z^2}\sum_{n=2}^\infty \binom{\frac12}{n}(-4z)^n\\
&=\frac{1}{4z^2}\left(1-2z-\sum_{n=0}^\infty \binom{\frac12}{n}(-4z)^n\right)\\
&=\frac{1}{4z^2}\left(1-2z-\sqrt{1-4z}\right)
\end{align*}$$
where the formula for the binomial series was used.
$f\left(-\frac1{16}\right)$ evaluates to $72-32\sqrt{5}$, and replacing the series with this value and simplifying finally yields $\phi$.
</s>