A Series For the Golden Ratio 
Question: Can we show that $$\phi=\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2} $$; where $\phi={1+\sqrt{5} \above 1.5pt 2}$ is the golden ratio ?


Some background and motivation: 
Wikipedia only provides one series for the golden ratio - see also the link in the comment by @Zacky. I became curious if I could construct another series for the Golden Ratio based on a slight modification to a known series representation of the $\sqrt{2}.$ At first I considered $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{(2n)!!}$$; which series can be accelerated via an Euler transform to yield $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{2^{3n+1}(n!)^2}$$ This last series became the impetus to try and and get to the Golden ratio. Through trial and error I stumbled upon
$$\frac{\sqrt{5}}{11}=\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}$$

Using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
$$\dfrac{(2n)!}{5^{3n+1}(n!)^2}=\dfrac{2^n\cdot1\cdot3\cdot5\cdots(2n-3)(2n-1)}{5^{3n+1}n!}=\dfrac{-\dfrac12\left(-\dfrac12-1\right)\cdots\left(-\dfrac12-(n-1)\right)}{n!\cdot5}\left(-\dfrac4{5^3}\right)^n$$
$$\implies5\sum_{n=0}^\infty\dfrac{(2n)!}{5^{3n+1}(n!)^2}=\left(1-\dfrac4{5^3}\right)^{-1/2}=?$$
Alternatively using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
Using Generalized Binomial Expansion, $$(1+x)^m=1+mx+\frac{m(m-1)}{2!}x^2+\frac{m(m-1)(m-2)}{3!}x^3+\cdots$$ given the converge holds
$mx=\dfrac{2!}{5^3(1!)^2},\dfrac{m(m-1)}2x^2=\dfrac{4!}{5^6(2!)^2}$
$\implies m=-\dfrac12,x=-\dfrac4{5^3}$
