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For fun, I have been considering the number

$$ \ell := \sum_{p} \frac{1}{2^p} $$

It is clear that the sum converges and hence $\ell$ is finite. $\ell$ also has the binary expansion $$ \ell = 0.01101010001\dots_2 $$ with a $1$ in the $p^{th}$ place and zeroes elsewhere. I have also computed a few terms (and with the help of Wolfram Alpha, Plouffe's Inverter, and this link from Plouffe's Inverter) I have found that $\ell$ has the decimal expansion

$$ \ell = .4146825098511116602481096221543077083657742381379169778682454144\dots. $$ Based on the decimal expansion and the fact that $\ell$ can be well approximated by rationals, it seems exceedingly likely that $\ell$ is irrational. However, I have been unable to prove this.

Question: Can anyone provide a proof that $\ell$ is irrational?

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    And how to prove it is irrational if the sum is taken over any infinite subset of primes? Sorry i misread, would be more interesting if you asked about transcendental.2012-02-16
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    Perhaps the question you meant to ask was whether $\ell$ is _trancendental_?2012-02-16
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    I thought about it being transcendental or algebraic also, but I did not see how first to prove that it was irrational. It seems I overlooked the obvious here. At any rate, is it also easy to see if $\ell$ is transcendental?2012-02-16
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    The sequence of digits appears, with some references, at https://oeis.org/A0510062012-02-16
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    It is *never* easy to see that a number is transcendental (aside from Liouville numbers).2012-02-17
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    @JavaMan: It is certainly not easy for me! Presumably your number is transcendental, "almost all" real numbers are. But the gaps between consecutive primes are not even close to being large enough to use a Liouville-type approximatibility argument.2012-02-17
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    Also discussed here: http://mathoverflow.net/questions/24270/a-number-encoding-all-primes2012-02-17
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    I proved explicitly a few weeks ago that the sequence of prime numbers is not periodic. The proof is trivial see: http://arxiv.org/abs/1305.0954. Hopefully of use here also.2013-12-06

3 Answers 3

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That $\ell$ is irrational is clear. There are arbitrarily large gaps between consecutive primes, so the binary expansion of $\ell$ cannot be periodic. Any rational has a periodic binary expansion.

The fact that there are arbitrarily large gaps between consecutive primes comes from observing that if $n>1$, then all of $n!+2, n!+3, \dots, n!+n$ are composite.

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If it was rational, then the binary expansion would eventually repeat -- but the distribution of primes doesn't repeat.

To wit: if the repeat had period $n$ and $p$ was a prime large enough to be inside the repeating part, then $p+pn=(1+n)p$ would have to be prime, which obviously isn't the case. On the other hand, the repeating period cannot consist of all zeroes, because there are infinitely many primes.

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    Thanks for your help. I wish I could've accepted both answers.2012-02-17
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Well to prove that this number is irrational you must know that a rational number has a periodic sequence of digits in any basis after a fixed digit. That is if $r=0.f_{1}f_{2}...f_{n}...$ then $f(n+T)=f(n)$ for some $T \in \mathbb{N}$ and $\forall n \geq k_{0} \in \mathbb{N}$. (check that!).

Then supposing by absurd, let $n_{0}$ be a natural such that $n_0\geq k_0$ and $n_{0}$ is prime then $1=f(n_{0})=f(n_{0}+T)=f(n_{0}+2T)=...=f(n_{0}+n_{0}T)=1$ absurd because $n_{0}+n_{0}T$ is not prime.

An interesting question arises is that number Algebraic, i.e., that is solution of a polinomial equation with rational coefficients?