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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. –  user1708 Feb 16 '12 at 23:49
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Perhaps the question you meant to ask was whether $\ell$ is trancendental? –  Henning Makholm Feb 16 '12 at 23:55
    
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? –  JavaMan Feb 16 '12 at 23:57
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The sequence of digits appears, with some references, at oeis.org/A051006 –  Gerry Myerson Feb 16 '12 at 23:59
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Also discussed here: mathoverflow.net/questions/24270/a-number-encoding-all-primes –  user1708 Feb 17 '12 at 0:07

3 Answers 3

up vote 25 down vote accepted

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. –  JavaMan Feb 17 '12 at 0:19

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?

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