On the sums of primes

Question

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: $$\sum_{k=0}^np_{2k+1} \sim f(n)$$

Answer 1

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?