Looking for a limit

The answer for the continuous version when the binomials are replaced with the $\Gamma$ function is $$\alpha = \frac{\log\left(-\frac{\log(1+a)}{\log(1-a)}\right)}{\log \frac{1-a}{1+a}}$$ and when the binomials are replaced with a normal distribution of mean $K/2$ and variance $K/4$. $$\alpha = -2\frac{\log \left(1-a^2\right)}{\log ^2\left(\frac{1-a}{a+1}\right)}$$

The discrete version does not converge. Consider $a=\phi^{-1}$ (the golden ratio conjugate) and $x=(2\phi+1)^n$ for some $n \in \mathbb{N}$, the variance will be equal to $x$ for $j=n+2K/3$. The tail exponent oscillates with a period of 3 and does not converge as $K\rightarrow \infty$

However, for all intent and purposes, the first $\alpha$ given is a good description of the asymptotic behavior of the tail, the tail just isn't smooth enough to see it at the infinitesimal scale.