Golden series

My view using generating function of Central binomial coefficients. $S=\frac{1}{64}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(2n+1)!}{16^n n!(n+1)!}=\frac{1}{64}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(2n+1)}{16^n (n+1)(n+2)}{ 2n\choose n}\\=\frac{1}{64}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{16^n}\left(\frac{3}{n+2}-\frac{1}{n+1}\right){2n\choose n}$ Recall that the generating function for central binomial coefficients and Catalan number ( which can be easily derived from CBC generating function). For $|x|<\frac{1}{4}$ $\sum_{n=0}^{\infty}{2n\choose n}x^n=\frac{1}{\sqrt{1-4x}}\cdots (1),\; \; \sum_{n=0}^{\infty}\frac{1}{n+1}{2n\choose n}x^n=\frac{2}{1+\sqrt{1-4x}}\cdots(2)$ set $x=-\frac{1}{16}$ in $(2)$ , we get then $-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{16^n(n+1)}{2n\choose n}=-\frac{4}{2+\sqrt 5}=8-4\sqrt 5$ Similarly multiple by $x$ in equation first and then integrate it from $x=0$ to $x=-\frac{1}{16}$ . $-\sum_{n=0}^{\infty}\frac{(-1)^{n+2}}{16^n(n+2)}{2n\choose n}=-\int_0^{-\frac{1}{16}}\frac{256x}{\sqrt{1-4x}}dx=\frac{4}{3}\left(7\sqrt 5 -16\right)$ and hence $\frac{3}{18}+S= \frac{3}{18}+\frac{1}{64}\left(28\sqrt 5 -64-8+4\sqrt 5\right)=\frac{1+\sqrt 5}{2}=\phi$ giving us $\phi^2-\phi=(\phi+1)-\phi=1$ .

$\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n+1}\left(2n+1\right)!}{4^{2n+3}n!\left(n+2\right)!}$ $=\frac{1}{4}\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{16}\right)^{n+1}\left(2n+2\right)!}{2\left(n+2\right)\left(n+1\right)!^{2}}$ $=\frac{1}{8}\sum_{n=0}^{\infty}\frac{\left(-\frac{1}{16}\right)^{n+1}}{n+2}{2n+2\choose n+1}$ $=-2\sum_{n=1}^{\infty}\frac{\left(-\frac{1}{16}\right)^{n+1}}{n+1}{2n\choose n}$ $=-2\int_{0}^{-\frac{1}{16}}\left(\sum_{n=0}^{\infty}t^{n}{2n\choose n}-1\right)dt$ $=-2\int_{0}^{-\frac{1}{16}}\left(\frac{1}{\sqrt{1-4t}}-1\right)dt$ $=-2\left[\frac{-1}{2}\sqrt{1-4t}-t\right]_{0}^{-\frac{1}{16}}$ $=\frac{\sqrt{5}}{2}-\frac{9}{8}$

$\frac{13}{8}+\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n+1}\left(2n+1\right)!}{4^{2n+3}n!\left(n+2\right)!}$ $=\frac{13}{8}+\frac{\sqrt{5}}{2}-\frac{9}{8}$ $=\frac{1+\sqrt{5}}{2}=\varphi$

$\varphi^{2}-\varphi=\boxed{1}$

$\begin{aligned} \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n+1 }\left( 2n+1 \right) ! }{ { 4 }^{ 2n+3 }n!\left( n+2 \right) ! } } =&\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n+1 }\Gamma \left( 2n+2 \right) }{ { 4 }^{ 2n+3 }\Gamma \left( n+1 \right) \Gamma \left( n+3 \right) } } &&&&&&&&&&&& \textcolor{#20A900}{\textrm{(Duplication formula)} \quad\Gamma \left( 2n+2 \right) =\frac { { 2 }^{ 2n+1 } }{ \sqrt { \pi } } \Gamma \left( n+\frac { 3 }{ 2 } \right) \Gamma \left( n+1 \right) } \\=&\frac { 1 }{ \sqrt { \pi } } \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n+1 }{ 2 }^{ 2n+1 }\Gamma \left( n+\frac { 3 }{ 2 } \right) \Gamma \left( n+1 \right) }{ { 4 }^{ 2n+3 }\Gamma \left( n+1 \right) \Gamma \left( n+3 \right)} }&&&&&&&&&&&& \textcolor{#20A900}{\textrm{Multiply both numerator and denominator by }\Gamma \left( \frac { 3 }{ 2 } \right) } \\=&\frac { 1 }{ \sqrt { \pi } } \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n+1 }\Gamma \left( n+\frac { 3 }{ 2 } \right) \Gamma \left( \frac { 3 }{ 2 } \right) }{ 2^{ 2n+5 }\Gamma \left( n+3 \right) \Gamma \left( \frac { 3 }{ 2 } \right) } } &&&&&&&&&&&&\textcolor{#BA33D6}{ \frac { \Gamma \left( x \right) \Gamma \left( y \right) }{ \Gamma \left( x+y \right) } =B\left( x;y \right)} \\=&\frac { 1 }{ \Gamma \left( \frac { 3 }{ 2 } \right) \sqrt { \pi } } \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n+1 }B\left( n+\frac { 3 }{ 2 } ;\frac { 3 }{ 2 } \right) }{ 2^{ 2n+5 } } } &&&&&&&&&&&& \textcolor{#BA33D6} {\textrm{Beta function}\quad B\left( x;y \right) =\int _{ 0 }^{ 1 }{ { t }^{ x-1 }{ \left( 1-t \right) }^{ y-1 }dt } } \\=&\frac { 1 }{ { 2 }^{ 5 }\Gamma \left( \frac { 3 }{ 2 } \right) \sqrt { \pi } } \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n+1 } }{ 2^{ 2n } } \int _{ 0 }^{ 1 }{ { t }^{ n+\frac { 1 }{ 2 } }{ \left( 1-t \right) }^{ \frac { 1 }{ 2 } }dt } }\\=&\frac { \left( -1 \right) }{ { 2 }^{ 5 }\Gamma \left( \frac { 3 }{ 2 } \right) \sqrt { \pi } } \int _{ 0 }^{ 1 }{ { \left( 1-t \right) }^{ \frac { 1 }{ 2 } } } { t }^{ \frac { 1 }{ 2 } }\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }{ t }^{ n } }{ { 4 }^{ n } } dt } &&&&&&&&&&&& \textcolor{#BA33D6}{\textrm{Geometric series}\quad\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }{ t }^{ n } }{ { 4 }^{ n } } =\frac { 4 }{ t+4 } } }\\=&\frac { \left( -4 \right) }{ { 2 }^{ 5 }\Gamma \left( \frac { 3 }{ 2 } \right) \sqrt { \pi } } \int _{ 0 }^{ 1 } \frac { \sqrt { t\left( 1-t \right) } }{ 4+t } dt &&&&&&&&&&&& \textcolor{#20A900}{\textrm{By trig and U substitution} \int _{ 0 }^{ 1 } \frac { \sqrt { t\left( 1-t \right) } }{ 4+t } dt=\left( \frac { 9-4\sqrt { 5 } }{ 2 } \right) \pi }\\=&\frac { \left( -4 \right) }{ { 2 }^{ 5 }\Gamma \left( \frac { 3 }{ 2 } \right) \sqrt { \pi } } \left( \frac { 9-4\sqrt { 5 } }{ 2 } \right) \pi &&&&&&&&&&&& \textcolor{#20A900}{\textrm{substitute}\quad\Gamma \left( \frac { 3 }{ 2 } \right) =\frac { \sqrt { \pi } }{ 2 } }\\=&\frac { 4\sqrt { 5 } -9 }{ 8 } \end{aligned}$ $A=\frac { 13 }{ 8 } +\frac { 4\sqrt { 5 } -9 }{ 8 } =\frac { \sqrt { 5 } +1 }{ 2 } =\varphi(\textcolor{#CEBB00}{\textrm{Golden Ratio}} )$ $\textrm{Therefore we get}$ $\boxed{{ { \varphi } }^{ 2 }-\varphi =1}$