For $0 \leq s < 1$, consider
$$ \sum_{k=0}^\infty \sum_{l=0}^\infty s^{k-1} (1-s^l) \, k(l-k) \, p_k \, p_{l}, $$ where $(p_k)_{k \geq 0 }$ denotes a probability distribution with finite expectation, that is $$ \sum_{k=0}^\infty p_k = 1 \text{ and } \sum_{k=0}^\infty k \, p_k < \infty .$$
I want to show that
$$ \sum_{k=0}^\infty \sum_{l=0}^\infty s^{k-1} (1-s^l) \, k(l-k) \, p_k \, p_{l} \geq 0 $$ for all $0 \leq s < 1$.
Edit: Before, the question asked was:
I have difficulties showing that this double sum is nonnegative for $0 \leq s < 1$:
$$ \sum_{k=0}^\infty \sum_{l=0}^\infty s^{k-1} (1-s^l) \, k(l-k) \geq 0.$$
Note that the sums are absolutely converging. I have tried writing
$$\sum_{k=0}^\infty \sum_{l=0}^\infty s^{k-1} (1-s^l) \, k(l-k) = \sum_{l=0}^\infty \sum_{k=0}^l s^{k-1} (1-s^{l-k}) \, k (l-2k), $$
but I don't know whether this helps.
It was pointed out that the sum does not converge and I therefore modified the question and added the factors $p_k$.
In our case $\sum\limits_{k=0}^\infty\sum\limits_{l=0}^\infty a_{k,l}=\sum\limits_{0<k<l}(a_{k,l}+a_{l,k})$ since $a_{k,l}=0$ if $k=0$ or $l=0$ or $k=l$. So, our sum is $$\sum_{0<k<l}p_k p_l(l-k)\big(ks^{k-1}(1-s^l)-ls^{l-1}(1-s^k)\big)=\sum_{0<k<l}p_k p_l(l-k)ks^{k-1}\big(1-f_{l/k}(s^k)\big)\color{blue}{\geqslant 0},$$ where $f_a(x)=ax^{a-1}-(a-1)x^a$ grows from $f_a(0)=0$ to $f_a(1)\color{blue}{=1}$ when $a>1$, since $$f_a'(x)=a(a-1)x^{a-2}(1-x)>0\quad\text{when}\quad x\in(0,1).$$