It's kinda weird summation - 2

Consider the binomial series expansion of $\sqrt{1+x}$ as follows:

$\begin{aligned} \sqrt{1+x} & = 1 + \frac x2 - \frac {x^2} {2^2\cdot 2!} + \frac {1\cdot 3x^3}{2^3\cdot 3!} - \frac {1\cdot 3\cdot 5 x^4}{2^4\cdot 4!} + \cdots \\ & = 1 + \frac x2 - \sum_{n=0}^\infty \frac {(-1)^n{\color{#3D99F6}(2n+1)!!}x^{n+2}}{2^{n+2}(n+2)!} & \small \color{#3D99F6} \text{Note that }(2n+1)!! = \frac {(2n+1)!}{2^n n!} \\ & = 1 + \frac x2 - \sum_{n=0}^\infty \frac {(-1)^n(2n+1)!x^{n+2}}{4^{n+1}n!(n+2)!} & \small \color{#3D99F6} \text{Putting }x = \frac 14 \\ \sqrt{1+\frac 14} & = 1 + \frac 18 - \color{#3D99F6} \sum_{n=0}^\infty \frac {(-1)^n(2n+1)}{4^{2n+3}n!(n+2)!} \\ \frac {\sqrt 5}2 & = 1 + \frac 18 - \color{#3D99F6} S \\ \implies S & = 1 + \frac 18 - \frac {\sqrt 5}2 = \frac {9-4\sqrt 5}8 \end{aligned}$

Therefore, $A+B+C+D = 9+4+5+8 = \boxed{26}$ .