A well-known result is that $\sum 1/n^s$ converges for $\operatorname{Re}(s)>1$.
Question: For which $s$ does $\sum 1/p^s$ converge, where $p$ is over all primes?
Notes:
- Intuitively there are fewer $p$ than $n$ and so the sum might converge over a larger domain, perhaps $s>0.5$?
- We know $\sum 1/p$, where $s=1$, diverges. So by Dirichlet theory, the abscissa of convergence is $s>1$. Is this correct?
If the abscissa of convergence of a general Dirichlet series of the form $\sum_{n=1}^\infty a_n e^{-\lambda_n s}$ is non-negative, it is given by $$\limsup_{n\to\infty}\frac{\log|a_1+a_2+\cdots+a_n|}{\lambda_n}.$$ Here $a_n,s\in\mathbb C$ and $\{\lambda_n\}$ is a strictly increasing sequence of nonnegative real numbers that tends to infinity. The formula for the abscissa of convergence, which is stated above, is proven in this paper in theorem $5$.
Our series over the primes is a special case of the general Dirichlet series. Let $a_n=1$ and $\lambda_n=\log p_n$, so that $\sum_{n=1}\frac1{p_n^s}=\sum_{n=1}^\infty 1\cdot e^{-\log (p_n) s}$. The abcissa of convergence can be calculated with the asymptotic formula $p_n\sim n\log n$, which is proven here, as $$\limsup_{n\to\infty}\frac{\log n}{\log p_n}=\lim_{n\to\infty}\frac{\log n}{\log n+\log\log n}=1.$$
For further properties, you can look at this Wikipedia article of the Prime zeta function.