Factorial by Factorial

$\begin{aligned} S & = \sum_{n=0}^\infty \frac {(2n)!}{(2n+5)!} \\ & = \sum_{n=0}^\infty \frac 1{(2n+1)(2n+2)(2n+3)(2n+4)(2n+5)} & \small \blue{\text{By partial fraction decomposition}} \\ & = \frac 1{24} \sum_{n=0}^\infty \left(\frac 1{2n+1} - \frac 4{2n+2} + \frac 6{2n+3} - \frac 4{2n+4} + \frac 1{2n+5} \right) \end{aligned}$

$\begin{aligned} \ \ \ & = \frac 1{24} \left(\blue{\frac 11 - \frac 42 + \frac 63 - \frac 44 + \frac 15} + \red{\frac 13 - \frac 44 + \frac 65 - \frac 46 + \frac 17} + \blue{\frac 15 - \frac 46 + \frac 67 - \frac 48 + \frac 19} + \cdots \right) \\ & = \frac 1{24} \left(\frac 11 - \frac 42 + \frac 73 - \frac 84 + \frac 85 - \frac 86 + \frac 87 - \frac 88 + \frac 89 - \cdots \right) \\ & = \frac 1{24} \left(\frac 81 - \frac 82 + \frac 83 - \frac 84 + \frac 85 - \frac 86 + \frac 87 - \frac 88 + \frac 89 - \cdots \right) - \frac 1{24} \left(\frac 71 - \frac 42 + \frac 13 \right) \\ & = \frac 13 \ln 2 - \frac 29 \end{aligned}$

Therefore $\dfrac 1{A+B} = \dfrac 1{\frac 13-\frac 29} = \boxed 9$ .

$\large \displaystyle S = \sum_{n=0}^{\infty} \frac{ (2n)!}{(2n+5)!}$

$\large \displaystyle S = \sum_{n=0}^{\infty} \frac{ (2n)!}{(2n)! \cdot (2n+1)(2n+2)(2n+3)(2n+4)(2n+5)}$

$\large \displaystyle S = \sum_{n=0}^{\infty} \frac{ 1}{ (2n+1)(2n+2)(2n+3)(2n+4)(2n+5)}$

By partial fractions:

$\large \displaystyle S = \frac{1}{24} \left [ \sum_{n=0}^{\infty} \frac{1}{2n+1} - \frac{4}{2n+2} + \sum_{n=0}^{\infty} \frac{6}{2n+3} - \frac{4}{2n+4} + \sum_{n=0}^{\infty} \frac{1}{2n+5} \right ]$

On second sum, make $k = n+1$ . On third sum, make $l = n+2$ :

$\large \displaystyle S = \frac{1}{24} \left [ \sum_{n=0}^{\infty} \frac{1}{2n+1} - \frac{4}{2n+2} + \sum_{k=1}^{\infty} \frac{6}{2k+1} - \frac{4}{2k+2} + \sum_{l=2}^{\infty} \frac{1}{2l+1} \right ]$

Taking off the first two terms of the first sum, the first term of the second sum, and standardizing the variable for all sums as $n$ :

$\large \displaystyle S = \frac{1}{24} \left [ 1 - 2 + \frac13 - 1 + \sum_{n=2}^{\infty} \frac{1}{2n+1} - \frac{4}{2n+2} + 2 - 1 + \sum_{n=2}^{\infty} \frac{6}{2n+1} - \frac{4}{2n+2} + \sum_{n=2}^{\infty} \frac{1}{2n+1} \right ]$

$\large \displaystyle S = \frac{1}{24} \left [ -\frac23 + \sum_{n=2}^{\infty} \frac{8}{2n+1} - \frac{8}{2n+2} \right ]$

$\large \displaystyle S = \frac{1}{24} \left [ -\frac23 + 8\sum_{n=5}^{\infty} \frac{ (-1)^{n+1}}{n} \right ]$

$\large \displaystyle S = \frac{1}{24} \left [ -\frac23 + 8 \left (\sum_{n=5}^{\infty} \frac{ (-1)^{n+1}}{n} \color{#3D99F6} + 1 - \frac12 + \frac13 - \frac 14 \color{#D61F06} - 1 + \frac12 - \frac 13 + \frac 14 \color{#333333} \right ) \right ]$

$\large \displaystyle S = \frac{1}{24} \left [ -\frac23 + 8 \left ( \sum_{ \color{#3D99F6}n=1}^{\color{#333333}\infty} \frac{ (-1)^{n+1}}{n} \color{#D61F06} - 1 + \frac12 - \frac 13 + \frac 14 \color{#333333} \right ) \right ]$

$\large \displaystyle S = \frac{1}{24} \left [ -\frac23 + 8 \left ( \sum_{n=1}^{\infty} \frac{ (-1)^{n+1}}{n} \right ) - \frac{14}{3} \right ]$

$\large \displaystyle S = \frac{1}{24} \left [ 8 \left ( \sum_{n=1}^{\infty} \frac{ (-1)^{n+1}}{n} \right ) -\frac{16}{3} \right ]$

$\large \displaystyle S = \frac{1}{3} \left ( \sum_{n=1}^{\infty} \frac{ (-1)^{n+1}}{n} \right ) - \frac{2}{9}$

From Maclaurin Series of $\ln(1+x) , |x| \leq 1, x \neq -1$ :

$\color{#20A900} \boxed{ \large \displaystyle S = \frac{1}{3} \ln(2)- \frac{2}{9}}$

So:

$\color{#3D99F6} \large \displaystyle A = \frac13, B = - \frac29 \rightarrow \boxed{\large \displaystyle \frac{1}{A+B} = 9}$