I'm working on finding power series of $\frac{1}{1 - z - 2z^2}$. And I found post about it here and The conclusion seems fair enough: $$\frac{1}{1-z-2 z^2} = \sum_{k=0}^{\infty} a_k z^k$$ $$a_k = \frac{2}{3} 2^k + \frac{1}{3} (-1)^k$$
But I'm wondering about the radius of convergence. I'm thinking $|z| < R$ with $R = \frac{1}{2}$ would be it since it is the smaller one between $1$ and $\frac{1}{2}$. Is this correct? And can I consider these values to be just like in real numbers? Also, if $|z| < R$ with $R = \frac{1}{2}$ is the right thing, what happens in other points? for example $$$$$|z|=\frac{1}{2}$ but not $z=\frac{1}{2}$ $$$$$\frac{1}{2} < |z| < 1$ $$$$$|z|=1$ but not $z=-1$ $$$$$1 < |z|$
Yes, the radius of convergence is $\frac12$. You can prove that using the fact that$$\frac1{1-z-2z^2}=\frac{1}{3 (z+1)}-\frac{2}{3 (2 z-1)}$$and expanding $\frac1{3(z+1)}$ and $\frac2{3(2z-1)}$ as power series; the first power series converges when $|z|<1$, whereas the second one converges when $|z|<\frac12$.