Is there an explicit asymptotic formula, in terms of $\alpha$, for the expression $$\displaystyle \sum_{p \leq x} \frac{1}{p^\alpha}$$ for $0 < \alpha < 1$?

The case $\alpha = 1$ is supplied by Merten's Theorem:

$$\displaystyle \sum_{p \leq x} \frac{1}{p} \sim \log \log x.$$

  • $\begingroup$ You can estimate the series by an integral like here $\endgroup$ Jan 27, 2015 at 1:21
  • $\begingroup$ $p_n\approx n\ln n$. $\endgroup$
    – Lucian
    Jan 27, 2015 at 1:24

1 Answer 1


The prime number theorem says that the $n$-th prime is $p_n \sim n \log n$. So your sum becomes $$ \sum_{p\le x}\frac{1}{p^{\alpha}} \sim \sum_{n : p_n \le x}\frac{1}{n^\alpha (\log n)^\alpha}\sim\sum_{n=2}^{x/ \log x}\frac{1}{n^\alpha (\log n)^\alpha} \sim \int_{2}^{x/\log x}\frac{dn}{n^{\alpha}(\log n)^\alpha} \sim \frac{n^{1-\alpha}}{(1-\alpha)(\log n)^{\alpha}}\Bigg\vert_{2}^{x/\log x}\sim\frac{x^{1-\alpha}}{(1-\alpha)(\log x)^{1-\alpha}(\log\log x)^{\alpha}} $$ for $\alpha \in (0,1)$.

  • 1
    $\begingroup$ $$ \frac{n^{1-\alpha}}{(1-\alpha)(\log n)^{\alpha}}\Bigg\vert_{2}^{x/\log x}\sim \frac{x^{1-\alpha}}{(1-\alpha)\log x}$$ $\endgroup$
    – reuns
    Mar 28, 2022 at 4:26

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