$\sum\limits_{\text{prime }p} 2^{-p}$ is an irrational number I need help to prove the following result.

$\displaystyle\sum_{\text{prime }p} 2^{-p}$ is an irrational number.

 A: One way to show that the sum is irrational
is the well-known fact that there
are arbitrarily long intervals
where there are no primes.
If the sum were rational,
chose the interval longer that the period,
and hello contradiction!
In particular,
$n!+k$ for $k = 2$ to $n$
has no primes, since each element is divisible
by a prime from $2$ to $n$.
This also shows that
$\sum_{\text{prime }p} m^{-p}$
is irrational for any integer $m \ge 2$.
A: Hint: Consider this number in base 2.  If it is rational, it must have a period.  What sort of consequence does this mean, that would lead to a contradiction?
A: If the $2$'s were replaced by $10$'s, the problem is clear. The decimal expansion of any rational number either terminates or repeats. The number in question is one that has $1$s at all prime places after the decimal point and zeros elsewhere. This does not terminate or repeat, so it must be irrational.
The problem you have is similar, except we consider the binary expansion. But the exact same argument works. 
