What's about $\sum_{k=1}^{n-1} p_{k} \sum_{l=k+1}^{n} p_{l}$ for prime numbers? By specialization of this formula, here in PROBLEMA 36, page 453 (in spanish), taking $\frac{1}{x_i}$ as the ith prime number we've (with at least two summands) 
$$ \left(  \sum_{k=1}^{n} p_{k}  \right)^{2} \geq\frac{2n}{n-1} \sum_{k=1}^{n-1} p_{k} \sum_{l=k+1}^{n} p_{l}, $$
and since 
$$ \sum_{k=1}^{n} p_{k}=p_n\pi(p_n)- \sum_{k=1}^{p_n-1} \pi(k) =np_n-\frac{1}{2}\frac{p^2_n}{\log p_n} +O \left( \frac{p_n^2}{\log^2p_n} \right) $$
then 
$$\sum_{k=1}^{n-1} p_{k} \sum_{l=k+1}^{n} p_{l}=\frac{n-1}{2n} \left( np_n-\frac{1}{2}\frac{p_n^2}{\log p_n} +O \left( \frac{p_n^2}{\log^2p_n} \right) \right)^{2}. $$

Question. Is it right? Can you improve it or do a right simplification? If you can improve the computations for the behaviour of $$\sum_{k=1}^{n-1} p_{k} \sum_{l=k+1}^{n} p_{l}$$
  please show us your approach. Thanks in advance.

My attempt is think that I can compute for $n\geq 6$ with the known inequlity
$$n(\log(n\log n)-1)<p_n<n\log(n\log n)$$ then using the first inequality $\log(n(\log(n\log n)-1))<\log(p_n)$ thus
 $$\frac{1}{\log p_n}<\frac{1}{\log(n(\log(n\log n)-1))},$$ also I need to use the second inequality, and binomial theorem to get the main term and the error term. 
 A: Arguing very naively, 
using
$p_k \approx k \ln(k)$
and
$\sum k^a \ln^b(k) 
\approx \int x^a \ln^b(x) dx
$ ,
$\begin{array}\\
\sum_{k=1}^{n-1} p_{k} \sum_{j=k+1}^{n} p_{j}
&\approx \sum_{k=1}^{n-1} k \ln(k) \sum_{j=k+1}^{n} j\ln(j)\\
&\approx \sum_{k=1}^{n-1} k \ln(k) \int_{x=k+1}^{n} x\ln(x)dx\\
&= \sum_{k=1}^{n-1} k \ln(k)  ( \frac12 x^2\ln(x)-\frac14 x^2)\big|_{k+1}^{n} \\
&= \sum_{k=1}^{n-1} k \ln(k)  ( \frac12 (n^2\ln(n)-k^2\ln(k))-\frac14 (n^2-k^2)) \\
&= (\frac12 n^2\ln(n)-\frac12 n^2) \sum_{k=1}^{n-1} k \ln(k) -\frac12\sum_{k=1}^{n-1} k^3\ln^2(k)+\frac14 \sum_{k=1}^{n-1} k^3 \ln(k) \\
&\approx (\frac12 n^2\ln(n)-\frac12 n^2) ( \frac12 n^2\ln(n)-\frac14 n^2) 
-\frac12( \frac14 n^4 \ln^2(x)-\frac18 n^2 \ln(x)+\frac1{32}n^4))
+\frac14 ( \frac14 n^4 \ln(n)-\frac1{16}n^4)) \\
&\approx \frac14 n^4\ln^2(n)-\frac38 n^4 \ln(n)+\frac18 n^4 
- \frac18 n^4 \ln^2(x)+\frac1{16} n^2 \ln(x)+\frac1{64}n^4
+ \frac1{16} n^4 \ln(n)-\frac1{64}n^4 \\
&= \frac18 n^4\ln^2(n)-\frac{5}{16} n^4 \ln(n)+\frac18 n^4 \\
&= \frac18 n^4(\ln^2(n)-\frac{5}{2}  \ln(n)+1) \\
\end{array}
$
I think there is a good chance
that first term is correct,
but I don't know about the others.
