Prime reciprocals sum Let $a_i$ be a sequence of $1$'s and $2$'s and $p_i$ the prime numbers.
And let $r=\displaystyle\sum_{i=1}^\infty p_i^{-a_i}$
Can $r$ be rational, and can r be any rational $> 1/2$ or any real?

ver.2:
Let $k$ be a positive real number and let $a_i$ be $1 +$ (the $i$'th digit in the binary decimal expansion of $k$).
And let $r(k)=\displaystyle\sum_{n=1}^\infty n^{-a_n}$
Does $r(k)=x$ have a solution for every $x>\pi^2/6$, and how many ?

The question with primes in the denominator:
The minimum that $r$ could possibly be is $C=\sum\limits_{i=1}^\infty\frac{1}{p_i^2}$. However, a sequence of $1$s and $2$s can be chosen so that $r$ can be any real number not less than $C$. Since $\sum\limits_{i=1}^\infty\left(\frac{1}{p_i}-\frac{1}{p_i^2}\right)$ diverges, consider the sum
$$
S_n=\sum_{i=1}^n b_i\left(\frac{1}{p_i}-\frac{1}{p_i^2}\right)
$$
where $b_i$ is $0$ or $1$.  Choose $b_n=1$ while $S_{n-1}+\frac{1}{p_n}-\frac{1}{p_n^2}\le L-C$ and $b_n=0$ while $S_{n-1}+\frac{1}{p_n}-\frac{1}{p_n^2}>L-C$.
If we let $a_i=1$ when $b_i=1$ and $a_i=2$ when $b_i=0$, then
$$
\sum\limits_{i=1}^\infty\frac{1}{p_i^{a_i}}=\sum\limits_{i=1}^\infty\frac{1}{p_i^2}+\sum_{i=1}^\infty b_i\left(\frac{1}{p_i}-\frac{1}{p_i^2}\right)=C+(L-C)=L
$$
The question with non-negative integers in the denominator:
Changing $p_n$ from the $n^{th}$ prime to $n$ simply allows us to specify $C=\frac{\pi^2}{6}$. The rest of the procedure follows through unchanged. That is, choose any $L\ge C$ and let
$$
S_n=\sum_{i=1}^n b_i\left(\frac{1}{i}-\frac{1}{i^2}\right)
$$
where $b_i$ is $0$ or $1$.  Choose $b_n=1$ while $S_{n-1}+\frac{1}{n}-\frac{1}{n^2}\le L-C$ and $b_n=0$ while $S_{n-1}+\frac{1}{n}-\frac{1}{n^2}>L-C$.
If we let $a_i=1$ when $b_i=1$ and $a_i=2$ when $b_i=0$, then
$$
\sum\limits_{n=1}^\infty\frac{1}{n^{a_i}}=\sum\limits_{n=1}^\infty\frac{1}{n^2}+\sum_{n=1}^\infty b_n\left(\frac{1}{n}-\frac{1}{n^2}\right)=C+(L-C)=L
$$
We don't need to worry about an infinite final sequence of $1$s in the binary number since that would map to a divergent series.
