# Irrationality of $\sum_{p\in\mathbb{P}} \frac{1}{2^{p}}$

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?
• To answer your question, "is $m$ irrational", Yes it is. To see this just prove the following proposition: Let $x$ be a rational number. Then the binary representation of $x$ is eventually periodic. Note that this is true for rationals in any base. – Baby Dragon Feb 17 '14 at 2:32
• Fot the irrationality measure, I make a vague conjecture. It is that the irrationality measure is high (the word high is what makes this vague). The reason for this is that the primes are sparse, so truncations of $m$ do a good job of approximating $m$ and so we have a high irrationality measure. I think that this certainly makes it highly intuitively plausible that $m$ is transcendental (although I do not have a formal proof to offer right now). – Baby Dragon Feb 17 '14 at 2:44
• It's not just irrational, but also transcendental. – Lucian Feb 17 '14 at 2:48
• @Lucian The gaps between primes are too small for the usual arguments relevant to Liouville numbers to apply. What variant of the argument do you have in mind? – Andrés E. Caicedo Feb 17 '14 at 3:57
• @BabyDragon: the primes are not nearly sparse enough to force the irrationality measure to be high. For example, the $k$th powers are far sparser than the primes, yet the truncations to the analogous series $\sum_{n=1}^\infty 2^{-n^k}$ are not even good enough to establish an irrationality measure greater than $1$. – Greg Martin Jan 4 '15 at 0:41