to define this sum you need a function $f(x)$
$f(x)=$ the sum of the prime powers of $x$
$12=3^1\times2^2$ so $f(12)=1+2$
$16=2^4$ so $f(16)=4$
my question does this sum converge and if so what does it converge to?
$$\frac{f(2)}1+\frac{f(3)}2+\frac{f(4)}4+\frac{f(5)}8+\frac{f(6)}{16}+\cdots+\frac{f(x)}{2^{x-2}}+\cdots$$
Well, it converges. You can estimate every $f(n)$ with $n$ itself, since $n \leq 2^n$ for every natural number. Then you have clearly convergence.
As to what it converges to, I don’t think it is possible to compute, and if it is I would be extremely surprised