$\varphi$ -tastical Summation!

Let's generalise the series

It is known from Generating function and Taylor series that

$\displaystyle\sum_{n=0}^∞ \dfrac{(-1)^n (2n)!}{2^{2n}(n!)^2} x^{2n} = \dfrac{1}{\sqrt{1+x^2}}$

Taking integral both sides from $0$ to $a$

$\sum_{n=0}^∞ \dfrac{(-1)^n (2n)! a^{2n+1}}{(2)^{2n}(2n+1).(n!)^2} =\int_0^a \dfrac{1}{\sqrt{1+x^2}} dx$ $\implies \sum_{n=0}^∞ \dfrac{(-1)^n (2n)! a^{2n+1}}{2^{2n}(2n+1).(n!)^2} =[\log(x+\sqrt{x^2+1})]_0^a$ $= \log(a+\sqrt{a^2+1} )$ Now put $a= \sqrt{\dfrac{2}{1+\sqrt{5}} }=\sqrt{\dfrac{1}{\phi }}$ The series becomes

$\displaystyle\sqrt{ \dfrac{2}{1+\sqrt{5}}}\sum_{n=0}^∞ \dfrac{(-1)^n (2n)!}{(2(1+\sqrt{5}))^n (2n+1)(n!)^2} = \log(\frac{1+\phi}{\sqrt{\phi}} )= \dfrac{3}{2} \log(\frac{1+\sqrt{5}}{2} )$

So $\dfrac{2}{3} \sum_{n=0}^∞ \dfrac{(-1)^n (2n)! }{(2n+1)(2(1+\sqrt{5}))^n (n!)^2 }= \sqrt{\dfrac{1+\sqrt{5}}{2}} \log(\dfrac{1+\sqrt{5}}{2} )$ Answer is $\boxed{A+B+C+D+E+F+G+H+I= 20}$