Rephrased

Repeatedly adding the digits of a number is the same as finding the digital root of a number, which is the same as the remainder after dividing that number by $9$ . Let $n = 9m + k$ for some integers $m$ and $k$ such that $0 \leq k < 9$ . Then the digital sum of $n$ is $k$ , and for $0 \leq k \leq 4$ the digital sum of $2n$ will be even and not in the form of $2p + 1$ (and therefore not applicable), but for $5 \leq k < 9$ the digital sum of $2n$ will be $2k - 9$ . Setting $2k - 9 = 2p + 1$ solves to $p = k - 5$ , therefore $a(k - 5) + b = k$ for $5 \leq k < 9$ . Then when $k = 5$ , $b = 5$ , and when $k = 6$ and $b = 5$ , $a = 1$ . Therefore $\frac{b + a}{b - a} = \frac{5 + 1}{5 - 1} = \frac{3}{2} = \boxed{1.5}$ .