But where to start from?

Start simplifying the ladder from the bottom: $\begin{aligned}\displaystyle\sum_{B=A}^{\infty}\dfrac{1}{B^2+3B+2}=&\displaystyle\sum_{B=A}^{\infty}\left(\dfrac{1}{B+1}-\dfrac{1}{B+2}\right)\\=&\dfrac1{A+1}~~~~(**)\end{aligned}$

$\displaystyle\sum_{A=1}^H 2(A+1)=H(H+3)$ $(\small{\text{Use }\displaystyle\sum r=\dfrac{n(n+1)}{2}})$

$\begin{aligned}\displaystyle\sum_{H=S}^{\infty}\dfrac{1}{H(H+3)}=&\displaystyle\sum_{H=S}^{\infty}\left(\dfrac{1}{H}-\dfrac{1}{H+3}\right)\\=&\dfrac1{S}+\dfrac{1}{S+1}+\dfrac1{S+2}~~~~(**)\end{aligned}$

$\displaystyle\sum_{S=1}^I 2(S)=I(I+1)$ $(\small{\text{Use }\displaystyle\sum r=\dfrac{n(n+1)}{2}})$

$\small{\text{Note:- }(**) \text{ stands for telescopic series.}}$

Now we are left with :-

$\large{\displaystyle\prod_{I=5}^{R}\dfrac{(I-3)(I+4)}{I(I+1)}\\=\displaystyle\prod_{I=5}^{R}\dfrac{I+4}{I+1}\times \displaystyle\prod_{I=5}^{R}\dfrac{I-3}{I}}$

(Both are $\mathbf{Telescopic~ Products}$ )

$=\dfrac{(R+2)(R+3)(R+4)}{6\cdot 7\cdot 8}\times \dfrac{2\cdot 3\cdot 4}{(R-2)(R-1)R}$

As $R\to \infty$ this product approaches $\dfrac{2\cdot 3\cdot 4}{6\cdot 7\cdot 8}$ i.e $\dfrac1{14}$ and we are asked its reciprocal which is $\large \boxed{\color{#007fff}{14}}$ .