The golden ratio
$$
x : 1 = 1 + x : x
$$
leads to the equation
$$
x^2 - x - 1 = 0 \quad (\#)
$$
It can be transformed to two different equations of the form
$$
x = F(x)
$$
which then can be used to substitute the $x$ on the right hand side by $F(x)$
$$
x = F(F(x)) = F(F(F(x))) = \cdots
$$
Your transformed version of $(\#)$ was this equation:
$$
x = 1 + \frac{1}{x} \quad (*)
$$
The repeated substitution of the RHS $x$ in $(*)$ with the RHS term
$$
x \mapsto 1 + \frac{1}{x}
$$
results in the expansion
\begin{align}
x
&= 1 + \frac{1}{x} \quad (\mbox{EX}1.1) \\
&= 1 + \frac{1}{1 + \frac{1}{x}} \quad (\mbox{EX}1.2) \\
&= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{x}}} \quad (\mbox{EX}1.3) \\
&= \cdots
\end{align}
and leads to the continued fraction
$$
x = 1 + \frac{1\rvert}{\lvert 1} + \frac{1\rvert}{\lvert1} + \cdots \quad (\mbox{CF}1)
$$
This continued fraction is positive (consisting only of additions and divisions of positive numbers) and will converge to the positive solution of $(\#)$.
Note that each of the steps $(\mbox{EX}1.n)$ is an equation equivalent
to equation $(\#)$, having two solutions for $x$, while $(\mbox{CF}1)$
converges only to the positive root.
This is because the continued fraction is the limit of these finite fractions:
\begin{align}
c_0 &= 1 \\
c_1 &= 1 + \frac{1}{1} = 2 \\
c_2 &= 1 + \frac{1}{1 + \frac{1}{1}} = 1.5 \\
c_3 &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}} = 1.\overline{6} \\
c_4 &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}}} = 1.6 \\
\end{align}
Note that compared to the expansions $(\mbox{EX}1.n)$ the right hand side $1/x$ is dropped, which makes the difference.
To get the other root, the golden ratio equation $(\#)$
needs to be transformed into
$$
x = \frac{1}{-1 + x} \quad (**)
$$
Repeated substitution of the RHS $x$ in $(**)$ with the RHS term
$$
x \mapsto \frac{1}{-1 + x}
$$
results in the expansion
\begin{align}
x
&= 0 + \frac{1}{-1 + x} \quad (\mbox{EX}2.1) \\
&= 0 + \frac{1}{-1 + \frac{1}{-1 + x}} \quad (\mbox{EX}2.2) \\
&= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1 + x}}} \quad (\mbox{EX}2.3) \\
&= \cdots
\end{align}
leads to the continued fraction
$$
x = 0 + \frac{1\rvert}{\lvert-1} + \frac{1\rvert}{\lvert-1} + \cdots \quad (\mbox{CF}2)
$$
for the negative root of $(\#)$.
Note that all equations $(\mbox{EX}2.m)$ are equivalent to equation $(\#)$ and the equations $(\mbox{EX}1.n)$, thus have two solutions for $x$.
However the derived continued fraction $(\mbox{CF}2)$ is the limit of these finite fractions:
\begin{align}
d_0 &= 0 \\
d_1 &= 0 + \frac{1}{-1} = -1 \\
d_2 &= 0 + \frac{1}{-1 + \frac{1}{-1}} = -0.5 \\
d_3 &= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1}}} = -0.\overline{6} \\
d_4 &= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1}}}} = -0.6\\
\end{align}
They result from dropping the right hand side $1/(-1+x)$ sub term.
Note that the negative root is nicely approached from above by the even numbered terms and from below by the odd numbered terms.
First of all note that
$$\frac1{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}n x^n$$
Lets rewrite your sum as the following
$$\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac15\sum_{n=0}^\infty\binom{2n}n\left(\frac1{5^3}\right)^n=\frac15\frac1{\sqrt{1-\left(\frac4{5^3}\right)}}=\frac{\sqrt 5}{11}$$
And therefore you can correctly conclude that
$$\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac12+\frac{11}2\frac{\sqrt 5}{11}=\frac{1+\sqrt 5}2$$
$$\therefore~\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\phi$$
Best Answer
Instead of representing $\frac{a}{b}$ as a fraction, represent it as the vector $\left( \begin{array}{c} a \\ b \end{array} \right)$.
Then, all you are doing to generate your sequence is repeatedly multiplying by the matrix $\left( \begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array} \right)$. One of the eigenvectors of this matrix is $\left( \begin{array}{c} \frac{\sqrt{5}-1}{2} \\ 1 \end{array} \right)$, which has a slope equal to the "golden ratio".
This is a standard example of a linear discrete dynamical system, and asymptotic convergence to an eigenvector is one of the typical things that can happen. You can also guess at the long-term behavior of the system by looking at its vector field.
https://kevinmehall.net/p/equationexplorer/#%5B-100,100,-100,100%5D&v%7C(x+y)i+(x+2y)j%7C0.1
In this case you see everything that starts in the first quadrant diverges to infinity along the path of the eigenvector I mentioned before. For your sequence, you started at $\left( \begin{array}{c} 2 \\ 5 \end{array} \right)$, which lies in the first quadrant.
Side note: There is nothing particularly special about the golden ratio, the matrix above, or the starting point of $\left( \begin{array}{c} 2 \\ 5 \end{array} \right)$ for this sequence. You can change the starting point to be in the negative quadrant if you want to diverge in the opposite direction, and you can change the matrix if you want to diverge along a differently sloped eigenvector.