It is known that: $$\sum_{k=1}^np_k \sim \frac12n^2\ln{n}$$ The proof of which I cannot find. Here is a related MSE post with an appreciable answer. Now, I wonder how one might go about finding $f(n)$ such that: $$\sum_{k=0}^np_{2k+1} \sim f(n)$$
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Let $A_n=\sum_{k=0}^np_{2k+1}$ and $B_n=\sum_{k=0}^np_{2k}$. Then $p_{2k}<p_{2k+1}<p_{2k+2}$ implies $$B_n<A_n< B_{n+1}$$ and therefore, since $\sum_{k=1}^np_k \sim \frac12n^2\ln{n}$, it follows $$2n^2\ln(n)\sim\sum_{k=1}^{2n+1}p_k= A_n+B_n< 2A_n< A_n+B_{n+1}=\sum_{k=1}^{2n+2}p_k\sim 2n^2\ln(n).$$ Hence $\sum_{k=0}^np_{2k+1}=A_n\sim n^2\ln(n)$. Moreover, we have also that $\sum_{k=0}^np_{2k}=B_n\sim n^2\ln(n)$.
P.S. As a reference for $\sum_{k=1}^np_k \sim \frac12n^2\ln(n)$ see Asymptotic expression for sum of first n prime numbers?
Robert Z
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Thank you; how did you obtain $2n^2\ln(n) \sim \sum_{k=1}^{2n+1}p_k$? – Descartes Before the Horse Jul 23 '19 at 19:16
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By your first line, $\sum_{k=1}^{2n+1}p_k \sim \frac12(2n+1)^2\ln(2n+1)\sim 2n^2\ln(n) $ – Robert Z Jul 23 '19 at 20:53
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What is a more accurate function asymptotic to $\ln(2n+1)$ than $\ln(n)$? I am not too familiar with big/little-oh notation, so it would be helpful to see how they appear in this function. Would it be $\ln(n) + o(\ln(2n+1))$? – Descartes Before the Horse Jul 24 '19 at 00:22
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We have that $\ln(2n+1)=\ln(n(2+1/n))=\ln(n)+\ln(2+1/n)=\ln(n)+O(1)\sim \ln(n)$. ($O(1)$ means that it is bounded) – Robert Z Jul 24 '19 at 05:35