I want to show that the radius of convergence of $$g(z) = \sum_{n=0}^{\infty} a_n z^n$$ is $1/2$ when $a_0 = 1$ and $a_n = 2a_{n-1} + 1$ for $n \geq 1$. Since $$g(z) = 1 + \sum_{n=1}^{\infty} a_n z^n = 1 + \sum_{n=1}^{\infty} (2a_{n-1} + 1)z^n = 2zg(z) + \frac{1}{1-z}$$ $$\iff g(z) = \frac{1}{(1-2z)(1-z)}, $$
I am wondering if I can somehow use this closed form formula for $g(z)$. Since $1-2z = 0$ for $z=1/2$, that makes me suspect that somehow that is related to the radius of convergence, but I haven't been able to figure out how... Can anyone help explain it to me?
You are nearly done: $$\frac{1}{(1-2z)(1-z)}=\frac{2}{1-2z}-\frac{1}{1-z} =\sum\limits_{n=0}2^{n+1}z^n-\sum\limits_{n=0}z^n=\\ \sum\limits_{n=0}(2^{n+1}-1)z^n$$ from infinite geometric progression, or $$a_n=2^{n+1}-1$$ This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=\frac{1}{2}$.