<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] This series appears in this place, referencing the derivation at "Golden Mean Series Derivation", both by Brian Roselle.  That derivation starts with $$\phi = \frac{\sqrt{5}+1}{2}, $$ replaces $\sqrt{5}$ with $f(x) = \sqrt{x},$ expands $f$ in a Taylor series about $4$ (having a radius of convergence $> 1$, although this is not stated), then evaluating that series at $5$.  Mr. Roselle describes finding the form of $f^{(a)}(x)$ via inspection.
We, however, can find these via induction.  We wish to find $f^{(a)}(4)$:


*

*$f^{(0)}(4) = 4^\frac{1}{2} = 2$.

*$f'(4) = \frac{1}{2} 4^\frac{-1}{2} = \frac{1}{4}$.

*$f''(4) = \frac{1}{2} \frac{-1}{2} 4^\frac{-3}{2} = \frac{-1}{32}$.

*Suppose $f^{(n)}(4) = \left. \frac{1}{2} \cdot\left( \frac{-1}{2} \cdot \frac{-3}{2} \cdot \cdots \cdot \frac{3-2n}{2} \right) x^{\frac{1}{2}-n} \right|_{x=4}$, which is certainly true for $n=2$.  Then $\left(f^{(n)}\right)'(4) = \left. \frac{1}{2} \cdot\left( \frac{-1}{2} \cdot \frac{-3}{2} \cdot \cdots \cdot \left( \frac{3}{2}-n \right) \cdot \left( \frac{3}{2}-(n+1) \right) \right) x^{\frac{1}{2}-(n+1)} \right|_{x=4} = f^{(n+1)}(4)$.


So induction has given us, for $n \geq 2$, 
$$\begin{align}
    f^{(n)}(4) &= \left. \frac{1}{2} \cdot\left( \frac{-1}{2} \cdot \frac{-3}{2} \cdot \cdots \cdot \frac{3-2n}{2} \right) x^{\frac{1}{2}-n} \right|_{x=4} \\
    &= 2^{-n}(-1)^{n-1}(2n-3)!! 4^{\frac{1}{2}-n}, 
\end{align}$$
where we have used $k!! = \begin{cases} k(k-2)(k-4) \cdots 4 \cdot 2, & k \text{ even}, \\ k(k-2)(k-4) \cdots 3 \cdot 1, & k \text{ odd} \end{cases}$.  Then for $x$ close enough to $4$ (and a quick ratio test shows $5$ is close enough), 
$$\begin{align}
    f(x) &= \sum_{n=0}^\infty \frac{f^{(n)}(4)}{n!}(x-4)^n \\
        &= \sum_{n=0}^\infty \frac{2^{-n}(-1)^{n-1}(2n-3)!! 4^{\frac{1}{2}-n}}{n!}(x-4)^n.
\end{align}$$
Replace $(2n-3)!!$ with $\frac{(2n-3)!}{2^{n-2}(n-2)!}$ and $x-4$ with $1$, pull out the first two summands and reindex $\sum_{n=2}^\infty \dots$ to start at $n=0$, plug into $\frac{f(5)+1}{2}$, and you get the given form of $\phi$.
</s>