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I am trying to find the following series:

$S=\displaystyle\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\dfrac{1}{p_ip_j},A\leq p_1 < p_2 < \dots < p_n \leq B, \lbrace A,B\rbrace \in \mathbb{N}$
Where $\lbrace p_1,p_2, \dots,p_n\rbrace$ are consecutive primes. I am stuck at $\displaystyle\sum_{i=1}^{n}\dfrac{1}{{p_i}^2}$.

This problem is related to checking the number of integers in a sequence that are divisible by two of the $n$ primes.

Example of $S$:
when $A = 10, B=20$, the primes are $\lbrace 11,13,17,19\rbrace$ and the sum $S$ is:
$\dfrac{1}{11\times 13}+\dfrac{1}{11\times 17}+\dfrac{1}{11\times 19}+\dfrac{1}{13\times 17} + \dfrac{1}{13\times 19}+\dfrac{1}{17\times 19}$

Work done thus far:
I have worked out that the sum is equivalent to:
$S=\dfrac{1}{2}\left( \left(\displaystyle\sum_{i=1}^{n}\dfrac{1}{p_i}\right)^2-\displaystyle\sum_{i=1}^{n}\dfrac{1}{{p_i}^2}\right)=\dfrac{1}{2}\left((S_1)^2-S_2\right)$
This can be done by doubling $S$ and adding $\displaystyle\sum_{i=1}^{n}\dfrac{1}{{p_i}^2}$ and realizing that it equals $\left(\displaystyle\sum_{i=1}^{n}\dfrac{1}{p_i}\right)^2$.
Say $S=\dfrac{1}{p_1p_2}+\dfrac{1}{p_1p_3}+\dfrac{1}{p_2p_3}$. Then $2S+\dfrac{1}{p_1p_1}+\dfrac{1}{p_2p_2} +\dfrac{1}{p_3p_3}= $
$\implies 2S=(S_1)^2+S_2$
This is sort of like completing a square matrix, where original sum is upper triangle, the doubling fills the lower triangle and the squared primes fills the diagonal.

Partial solution:
There is a nice estimation for $S_1:\displaystyle\sum_{p\in prime}^{x}=ln(ln(x))+B_1+o(1)$.
So given $A,B:$
$\approx \displaystyle\sum_{i=1}^{B}\dfrac{1}{p_i}-\displaystyle\sum_{i=1}^{A}\dfrac{1}{p_i}$
$\approx (ln(ln(B))+B_1+o(1))-(ln(ln(A))+B_1+o(1))$
$\approx ln(ln(B))-ln(ln(A))=ln(\dfrac{B}{A})$

Similarly, there is an estimation for the harmonic sum of prime squares.
However, this estimation apparently is for an infinite series and so I am unable to use a similar trick that derives $S_1$.

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