Vtyjkyuk

Let $a_i$ be a sequence of $1$'s and $2$'s and $p_i$ the prime numbers.

And let $r=displaystylesum_i=1^infty p_i^-a_i$

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 $r(k)=displaystylesum_n=1^infty n^-a_n$

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 $C=sumlimits_i=1^inftyfrac1p_i^2$. 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 $sumlimits_i=1^inftyleft(frac1p_i-frac1p_i^2right)$ diverges, consider the sum
$$
S_n=sum_i=1^n b_ileft(frac1p_i-frac1p_i^2right)
$$
where $b_i$ is $0$ or $1$. Choose $b_n=1$ while $S_n-1+frac1p_n-frac1p_n^2le L-C$ and $b_n=0$ while $S_n-1+frac1p_n-frac1p_n^2>L-C$.

If we let $a_i=1$ when $b_i=1$ and $a_i=2$ when $b_i=0$, then
$$
sumlimits_i=1^inftyfrac1p_i^a_i=sumlimits_i=1^inftyfrac1p_i^2+sum_i=1^infty b_ileft(frac1p_i-frac1p_i^2right)=C+(L-C)=L
$$
The question with non-negative integers in the denominator:

Changing $p_n$ from the $n^th$ prime to $n$ simply allows us to specify $C=fracpi^26$. The rest of the procedure follows through unchanged. That is, choose any $Lge C$ and let
$$
S_n=sum_i=1^n b_ileft(frac1i-frac1i^2right)
$$
where $b_i$ is $0$ or $1$. Choose $b_n=1$ while $S_n-1+frac1n-frac1n^2le L-C$ and $b_n=0$ while $S_n-1+frac1n-frac1n^2>L-C$.

If we let $a_i=1$ when $b_i=1$ and $a_i=2$ when $b_i=0$, then
$$
sumlimits_n=1^inftyfrac1n^a_i=sumlimits_n=1^inftyfrac1n^2+sum_n=1^infty b_nleft(frac1n-frac1n^2right)=C+(L-C)=L
$$
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.