Number Theory – Irrationality of Sum of 1/2^p Over All Primes

diophantine-approximationirrational-numbersnumber theoryprime numbers

Let $\mathbb{P}$ be the set of prime numbers, and consider $m=\displaystyle\sum_{p\in\mathbb{P}} \frac{1}{2^{p}}$. Is $m$ irrational?

In the following
paper, the author recalls several sufficient criteria for irrationality. When applying some of this criteria to $m$ I arrive that the condition of some of the criteria are not satisfied by means of Bertrand´s Postulate.

I found a similar result by Sándor:

Let $\lbrace a_m\rbrace$, $m\geq 1$ be a sequence of positive integers such that
$$\text{lim sup}\frac{a_{m+1}}{a_1a_2\cdots a_m}=\infty \;\;\;\;\text{and}\;\;\;\;\;\;\text{lim inf}\frac{a_{m+1}}{a_m}>1$$
Then the series $\displaystyle\sum_{m} \frac{1}{a_m}$ is an irrational number.

I have yet to prove that if $f$ is a continuous function then $\text{lim sup} \;f(a_m)=f(\text{lim sup}\, a_m)$. Assuming this;

$$\text{lim inf}\; P_{m+1}-P_{m}>1$$
where $P_m$ is the $m$-th prime.

But I have trouble with
$$\text{lim sup}\; P_{m+1}-\sum_{j=1}^{m}P_{j}$$
my guess is that
$$\text{lim sup}\; P_{m+1}-\sum_{j=1}^{m}P_{j}\neq\infty$$
so
$$\text{lim sup}\frac{2^{P_{m+1}}}{2^{P_1}\cdots 2^{P_m}}\neq\infty$$

and this theorem will result useless to tell if $m$ is irrational.

  1. How can I prove (or disprove) that $m$ is irrational (are there any other simpler criteria)?
  2. How can I use Dirichlet criterion or Hurwitz criterion?
  3. What is its irrationality measure?

Any help is highly appreciated.

Edit: That $m$ is irrational is clear by the criteria provided in the comments, namely that if $x$ is rational then the binary representation of $x$ is periodic. The number $m$ in binary expansion has $1$ in the $P$-th position and zero elsewhere. As there are arbitrarily large gaps between primes; then the binary representation fails to be periodic.

So the question that remains unsolved is:

  • What is its irrationality measure?

Best Answer

This number is called the prime constant. Since it is irrational because it doesn't have a periodic binary representation it's irrationality measure is 2 or more according to this article.

Again according to the article If the irrationality number of the prime constant were more than 2 it would be transcendental but as far as I can tell it is not known to be transcendental although it is listed as a suspected transcendental number, see:

http://en.wikipedia.org/wiki/List_of_numbers#Suspected_transcendentals.

So the prime number constant is irrational, its irrationality measure is 2 or more, and it is an open problem whether or not it is transcendental and whether or not its irrationality number is more than 2.

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