# Asymptotic formula for sums of powers of reciprocals of primes

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:

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

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

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 $\alpha \in (0,1)$.
• $$\frac{n^{1-\alpha}}{(1-\alpha)(\log n)^{\alpha}}\Bigg\vert_{2}^{x/\log x}\sim \frac{x^{1-\alpha}}{(1-\alpha)\log x}$$ Mar 28, 2022 at 4:26