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
$$\frac{\sqrt{5}}{11}=\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}$$
 A: 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
$mx=\dfrac{2!}{5^3(1!)^2},\dfrac{m(m-1)}2x^2=\dfrac{4!}{5^6(2!)^2}$
$\implies m=-\dfrac12,x=-\dfrac4{5^3}$
A: We know that the approximates by partial continued fractions of the golden ratios are the quotients of successive Fibonacci numbers $1,1,2,3,5,8,13,...$. Taking 
$$
\phi=\frac{1+\sqrt{5}}2\approx \frac{8}5\implies \sqrt5\approx \frac{11}{5}
$$
we get a lower approximation for $\sqrt5$. To the remainder of the square root after factoring out the approximation we can apply the the Newton/binomial series for the inverse square root
$$
\sqrt5=\frac{11}{5}\sqrt{\frac{125}{121}}=\frac{11}{5}\left(1-4\frac1{5^3}\right)^{-1/2}
=\frac{11}5\sum_{n=0}^\infty\binom{2n}{n}\frac1{5^{3n}}
$$
which is the series you got.

Exploring the same method for one place further in the Fibonacci sequence
$$
\phi=\frac{1+\sqrt{5}}2\approx \frac{13}8\implies \sqrt5\approx \frac{9}{4}
$$
gives an upper approximation of $\sqrt5$ and thus an alternating series,
$$
\sqrt5=\frac{9}{4}\sqrt{\frac{80}{81}}=\frac{9}{4}\left(1+4\frac1{320}\right)^{-1/2}
=\frac{9}4\sum_{n=0}^\infty\binom{2n}{n}\frac{(-1)^n}{320^{n}}
$$

Why that series?
The general binomial series reads as 
$$(1+x)^α=\sum_{n=0}^\infty\binom{α}n.$$ For $α=-1/2$ the binomial coefficient can be transformed as 
$$
\binom{-1/2}{n}=\frac{(-1/2)(-1/2-1)...(-1/2-n+1)}{n!}=(-1/2)^n\frac{(2n-1)(2n-3)...3\cdot 1}{n!}=(-1/4)^n\frac{(2n)!}{(n!)^2},
$$
resulting in the "simplified" formula 
$$\frac1{\sqrt{1-4x}}=\sum_{n=0}^\infty\binom{2n}nx^n.$$
A: 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
$$\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac12+\frac{11}2\frac{\sqrt 5}{11}=\frac{1+\sqrt 5}2$$

$$\therefore~\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\phi$$

