Anything known about the series $\sum_p \frac{1}{2^p} $, summing over the primes?

Question

Is anything known about this sum?

$$ \sum_{\text{prime}\;p} \frac{1}{2^p} $$

I've calculated that it converges, but I can't determine whether the following related sum can be analytically continued outside the unit circle:

$$\sum_{\text{prime}\;p} z^{p}$$

Answer 1

I will only address the second part of the question about $\sum\limits_{p \text{ prime}} z^p$.

There is a result by Pólya and Carlson.

Given any power series $f(z) = \sum\limits_{k=0}^\infty a_k z^k$ with integer coefficients and radius of convergence $1$, then $f(z)$ is either a rational function or has the unit circle $|z| = 1$ as natural boundary.

For the series at hand

$$f(z) = \sum\limits_{p \text{ prime}} z^p = \sum_{k=0}^\infty a_k z^k \quad\text{ where }\quad a_k = \begin{cases} 1, & p \text{ is prime}\\ 0, & \text{ otherwise }\end{cases}$$ It is trivial to see its radius of convergence is $1$. If $f(z)$ is rational, let's say $f(z) = \frac{p(z)}{q(z)}$ for polynomials $p(z)$ and $q(z)$, then $f(z)q(z) = p(z)$ is also polynomial.

WOLOG, assume $q(z)$ has the form $$q(z) = 1 + b_1 z + \cdots + b_n z^n \quad\text{ where } n = \deg(q)$$ Substitute this into the expression $f(z)q(z) = p(z)$ and compare coefficients of $z^{k+n}$ on both sides, we obtain

$$a_{k+n} + b_1 a_{k+n-1} + \cdots + b_n a_k = 0,\quad \text{ whenever }\quad k+n > \deg(p)$$

A consequence of this is if $a_k$ has infinitely many non-zero terms, then aside from initial $\deg(p)$ gaps, the gap between any two non-zero terms is at most $n-1$. It is known that there are infinitely many prime numbers and the prime gaps can be as large as one likes. This implies $f(z)$ is not rational.

By Pólya and Carlson, $\sum\limits_{p \text{ prime}} z^p$ has the unit circle as natural boundary. In other words, it cannot be analytic continued outside the unit disk in any manner.

Answer 2

The main thing to know is for $x > 0$ and by analytic continuation for $\Re(x) > 0$ $$f(x) = \sum_{p^k} \log(p) e^{-x p^k}, \qquad \Gamma(s) \frac{-\zeta'(s)}{\zeta(s)} = \int_0^\infty f(x)x^{s-1}dx$$

$$f(x) = \frac1{2i\pi} \int_{2-i\infty}^{2+i\infty}\Gamma(s) \frac{-\zeta'(s)}{\zeta(s)} x^{-s}ds = \sum Res(\Gamma(s) \frac{-\zeta'(s)}{\zeta(s)} x^{-s})$$ $$ = x^{-1}- \sum_\rho \Gamma(\rho) x^{-\rho}+\sum_{k=0}^\infty x^k(a_k+b_k \log(x))$$

$f$ is very interesting mainly because $f(x+2 i \pi \frac{a}{q})$ has a similar explicit formula in term of $\frac{L'(s,\chi)}{L(s,\chi)}$ for the Dirichlet characters modulo $q$, thus it encodes the generalized Riemann hypothesis.

$\sum_{p^k}\frac1k e^{-xp^k},\sum_p e^{-xp}$ work the same way except $\Gamma(s) \log\zeta(s),\Gamma(s)P(s)$ aren't meromorphic so their explicit formulas are more complicated.