Cfrgtkky

Question: Can we show that $$phi=frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2} $$; 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
$$frac{sqrt{5}}{11}=sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}$$

Note that

$$frac1{sqrt{1-4x}}=sum_{n=0}^{infty}binom{2n}n x^n$$

Lets rewrite your sum as the following to get

$$sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac15sum_{n=0}^inftybinom{2n}nleft(frac1{5^3}right)^n=frac15frac1{sqrt{1-left(frac4{5^3}right)}}=frac{sqrt 5}{11}$$

And therefore you can correctly concluded that

$$frac{1}{2}+frac{11}{2}sum_{n=0}^inftyfrac{(2n)!}{5^{3n+1}(n!)^2}=frac12+frac{11}2frac{sqrt 5}{11}=frac{1+sqrt 5}2$$

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

Using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+dots? $,

$$dfrac{(2n)!}{5^{3n+1}(n!)^2}=dfrac{2^ncdot1cdot3cdot5cdots(2n-3)(2n-1)}{5^{3n+1}n!}=dfrac{-dfrac12left(-dfrac12-1right)cdotsleft(-dfrac12-(n-1)right)}{n!cdot5}left(-dfrac4{5^3}right)^n$$

$$implies5sum_{n=0}^inftydfrac{(2n)!}{5^{3n+1}(n!)^2}=left(1-dfrac4{5^3}right)^{-1/2}=?$$

Alternatively using Calculating $1+frac13+frac{1cdot3}{3cdot6}+frac{1cdot3cdot5}{3cdot6cdot9}+frac{1cdot3cdot5cdot7}{3cdot6cdot9cdot12}+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

$mx=dfrac{2!}{5^3(1!)^2},dfrac{m(m-1)}2x^2=dfrac{4!}{5^6(2!)^2}$

$implies m=-dfrac12,x=-dfrac4{5^3}$

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
$$
phi=frac{1+sqrt{5}}2approx frac{8}5implies sqrt5approx frac{11}{5}
$$
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
$$
sqrt5=frac{11}{5}sqrt{frac{125}{121}}=frac{11}{5}left(1-4frac1{5^3}right)^{-1/2}
=frac{11}5sum_{n=0}^inftybinom{2n}{n}frac1{5^{3n}}
$$
which is the series you got.

Exploring the same method for one place further in the Fibonacci sequence
$$
phi=frac{1+sqrt{5}}2approx frac{13}8implies sqrt5approx frac{9}{4}
$$
gives an upper approximation of $sqrt5$ and thus an alternating series,
$$
sqrt5=frac{9}{4}sqrt{frac{80}{81}}=frac{9}{4}left(1+4frac1{320}right)^{-1/2}
=frac{9}4sum_{n=0}^inftybinom{2n}{n}frac{(-1)^n}{320^{n}}
$$

Why that series?

The general binomial series reads as
$$(1+x)^α=sum_{n=0}^inftybinom{α}n.$$ For $α=-1/2$ the binomial coefficient can be transformed as
$$
binom{-1/2}{n}=frac{(-1/2)(-1/2-1)...(-1/2-n+1)}{n!}=(-1/2)^nfrac{(2n-1)(2n-3)...3cdot 1}{n!}=(-1/4)^nfrac{(2n)!}{(n!)^2},
$$
resulting in the "simplified" formula
$$frac1{sqrt{1-4x}}=sum_{n=0}^inftybinom{2n}nx^n.$$