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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.

Community
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    write $\delta_P(n) = 1$ if $n$ is prime, $\delta_P(n) =0$ otherwise. consider the power series $P_p(x) = \sum_{n=0}^\infty \delta_P(n) x^n$ and the Dirichlet series $P_d(s) = \sum_{n=1}^\infty \delta_P(n) n^{-s}$. we know that $\ln \zeta(s) = \sum_{r=1}^\infty \frac{P_d(s r)}{r}$ hence $P_d(s) = \sum_n \frac{\mu(n)}{n} \ln \zeta(s) $ and from $n^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-nx} dx$ we get $\Gamma(s) P_d(s) = \int_0^\infty x^{s-1} \sum_{n=1}^\infty \delta_P(n) e^{-nx} dx = \int_0^\infty x^{s-1} P_p(e^{-x})dx = \ldots$ – reuns Mar 18 '16 at 22:01
  • I'm sorry this is stratospheric for me! I try read your comment tomorrow without squint, thanks! @user1952009 –  Mar 18 '16 at 22:06
  • There was a great mistake in the first step, the right statement is $$\left( \sum_{k=1}^{n} p_{k} \right)^{2} \geq\frac{2n}{n-1} \sum_{k=1}^{n-1} p_{k} \sum_{l=k+1}^{n} p_{l},$$ –  Mar 19 '16 at 07:22
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    $$2\sum_{k=1}^{n-1} p_{k} \sum_{\ell=k+1}^{n} p_\ell=\left(\sum_{k=1}^{n} p_{k}\right)^2- \sum_{k=1}^{n} p_k^2\sim\left(\frac12n^2\log n\right)^2$$ – Did Mar 20 '16 at 13:17
  • Very thanks much @Did, feel free to edit as an answer. –  Mar 21 '16 at 08:21
  • @Did: Comforting that your much simpler answer agrees with my complicated one. – marty cohen Mar 21 '16 at 19:37

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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.

marty cohen
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  • Very thanks much, also I have a great mistake when I take $\sqrt{}$ in LHS instead of $()^2$. I will read your answer, and wait other constributions this week. Very thanks much. –  Mar 19 '16 at 07:07
  • I have decided acceot your answer. I believe that it is enough for me, with your answer and Did's comment. Very thanks much for your good answer. –  Mar 21 '16 at 14:29