<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] Like 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? $,
$$T(n)=\frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}}=\dfrac1{64}\cdot\dfrac{1\cdot3\cdot5\cdots(2n+1)2^n}{(n+2)!16^n}$$
As the denominator has $(n+2)!,$
$$T(n)=-\dfrac1{8^{n+2}}\cdot\dfrac{-1\cdot1\cdot3\cdot5\cdots(2n+1)}{(n+2)!} $$ $$=\dfrac1{4^{n+2}}\cdot\dfrac{\dfrac12\left(\dfrac12-1\right)\left(\dfrac12-2\right)\cdots\left(\dfrac12-(n+1)\right)}{(n+2)!}$$
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
$$\sum_{n=0}^\infty T(n)=\left(1+\dfrac14\right)^{1/2}-1-T(-1)=\dfrac{\sqrt5}2-1-\dfrac14\dfrac12$$
See also: A Series For the Golden Ratio
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