I am trying to find a function of asymptotic growth of the Fibonacci sequence, which's generating function is $$ f(z)=\frac{1}{1-z-z^2}=\frac{1}{(1-\phi _-z)(1-\phi_+z)} $$ with poles at $ z_1=-\phi_- +2\pi ki $ and $ z_2=-\phi_++2\pi ki $ $$ where \space \phi_±=\frac{1±\sqrt{5}}{2}. $$ One of the necessary steps in finding the asymptotic growth of this sequence is to find the radius of the disk within which $f(z)$ is analytic, which would be easy if $f(z)$ had a pole at only one value of $z$ since radius would simply be the smallest magnitude of this pole in the imaginary plane.
But I am not sure what to do here since I have two poles giving me two different radii $r_1=|\phi_-|$ and $r_2=|\phi_+|$. Do I perhaps simply choose the radius with the smallest magnitude as I would if I had only one pole, or is there a more appropriate way to approach this problem?
Thank you for your time and help in advance.
The radius of convergence of a power series is the largest radius for which the series converges. In this case, the power series expansion of $f(z)$ is the Fibonacci sequence, so the radius of convergence of the Fibonacci sequence is the radius of convergence of the power series expansion of $f(z)$.
Since $f(z)$ has poles at $z_1=-\phi_- +2\pi ki$ and $z_2=-\phi_+ +2\pi ki$, the radius of convergence is the minimum of the magnitudes of these poles. This means that the radius of convergence is $r = \min(|\phi_-|, |\phi_+|) = |\phi_-| = \frac{1}{\sqrt{5}}$.
So the asymptotic growth of the Fibonacci sequence is governed by the singularity at $z=-\phi_-$, which has a magnitude of $\frac{1}{\sqrt{5}}$. This means that the Fibonacci sequence grows asymptotically like $\mathcal{O}\left(\frac{1}{\sqrt{5}}\right)$.