1

As a prospective undergraduate who has really benefited from his time on MSE thus far, i recently learnt that there exists asymptotic approximations for $\sum_{p\leq x} 1, \sum_{p\leq x} p, \sum_{p\leq x} \frac{1}{p}, \sum_{p\leq x} \frac{\log p}{p}, \sum_{p\leq x} \log p, \sum_{p^m\leq x, m\geq 1} \log p$ where $p$ is a prime, but have never seen any for $\sum_{p\leq x} p$ ?

Out of curiosity, i'm wondering if there is any such formula yet ?

An idea that quickly came to mind was, $\sum_{p\leq x} p$ = $\sum_{p\leq x} (p/\log p)\log p$, and then apply the Abel/Euler summation theorem ?

User1
  • 1,841

1 Answers1

1

$$\log \zeta(s-1) = \sum_{p^k} \frac{p^k}{k } p^{- sk}$$ its abscissa of convergence is $2$ and by the prime number theorem $\log \zeta(s-1) + \log(s-2)$ has a lower abscissa of convergence so $$\sum_{p^k \le x} \frac{p^k}{k}\sim \sum_{p \le x} p \sim \frac{x^2}{2\log x}$$

reuns
  • 77,999
  • I feel like your answer could be wrong: we have $${\sum_{k=1}^{n}p_{k}\sim\frac{n^{2}}{2}\log\left(n\right)}$$ (see here), and $\sum_{p \leq x} p = \sum_{k=1}^{\pi(x)} p_{k} \sim \frac{\pi(x)^{2}}{2}\log\left(\pi(x)\right) \sim \frac{x^{2}}{2 \ln(x)^2} ( \ln(x) - \ln \ln (x) ) \sim \frac{x^{2}}{2 \ln(x)}$ (since $\pi(x) \sim x/\ln(x)$ and logarithms of equivalent functions are equivalent). – Alphonse Jun 19 '17 at 19:18
  • @Alphonse yes I missed a $1/2$ coming from $\int_2^\infty \frac{x^{-s-2}}{2 \log x}dx\sim \log(s-2)$ – reuns Jun 19 '17 at 20:37