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

This question already has an answer here:

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

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

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

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• it blows up as $z \to 1$.... – mathworker21 Aug 24 at 16:18
• indeed mathworker – Ultradark Aug 24 at 16:19
• I guess you are asking if the sum is a lacunary function The primes don't satisfy the condition in the theorem of Hadamard given on the wiki, but it doesn't say that's a necessary condition. Maybe looking at the proofs of some of the theorems on lacunary functions will help. – saulspatz Aug 24 at 16:32
• It looks like this immediately satisfies the conditions of Fabry’s extension to Hadamard’s theorem: en.m.wikipedia.org/wiki/Fabry_gap_theorem – Erick Wong Aug 24 at 16:54

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.

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.

• @reuns can you explain this better? Maybe tailor your answer to my math level? – Ultradark Aug 24 at 16:30
• what do you not understand ? those functions have explicit formulas in term of the non-trivial zeros of $\zeta(s)$, this is of course a complicated subject, but the point is to understand that everything we know of those functions come from their explicit formula, thus we concentrate on $f$ whose explicit formula is easier to define. From that we know almost everything : analytic continuation, asymptotic... Which depends on the (G)RH. I don't think we know any closed-form/special value for $f(x)$ nor $\sum_p e^{-xp}$. – reuns Aug 24 at 16:32
• okay I understand better now :) – Ultradark Aug 24 at 16:37