Convergence radius of power series $\sum_{n=0}^\infty c_nz^n $ where $c_0 = 0$, $c_1 = 1$, $c_n =\frac{c_{n-1} + c_{n-2}}{2}$ In my homework I'm supposed to find the convergence radius of the power set in the title. I'm allowed to assume $C = \lim_{n\to\infty} {c_{n+1}\over c_n}$ exists and we should try to figure out that $C$ has to be.
I found $${c_{n+1}\over {c_n}} = {1\over 2} + {c_{n-1}\over{2c_n}}  $$
so for $n\to\infty$: $$C = {1\over2} + {1\over{2C}}  \Leftrightarrow C=1$$
which doesn't really help me further. Investigating $c_n - c_{n-1}$ didn't help me out either.
Any tips on what I might be missing here? I know how to find the radius once I've found the limit of the sequence, but can't seem to find it.
 A: Assume at the most general case for a sequence $c_n$ we have a recursive equation as following:$$c_n=\sum_{i=1}^{k}a_ic_{n-i}$$. Now define $C(z)=\sum_{n=0}^{\infty}c_nz^{-n}$. This transformation is also referred to as z-transform. By taking z-transform from both sides of the recursive equation we arrive at the following equation:$$C(z)=C(z)\sum_{i=1}^{k}a_iz^{-i}$$which leads to the $characteristic$ $equation$ as below:$$\sum_{n=1}^{k}a_n\lambda^{-n}=0$$
and the $c_n$ itself can be described as a linear combination of eigenfunctions $\lambda_i^n$ as following:$$c_n=\sum_{i=1}^{k}b_i\lambda_{i}^n$$where all the $\lambda_i$'s are assumed to be simple roots of the characteristic function.
Embarking on this, the characteristic equation of $2c_n=c_{n-1}+c_{n-2}$ is $2\lambda^2=\lambda+1$ which has two roots $\lambda=1$ and $\lambda=-\frac{1}{2}$ and leads us to $c_n=A(1)^n+B(-\frac{1}{2})^n$. Also $c_0=0$ and $c_1=1$ therefore $c_n=\frac{2}{3}-\frac{2}{3}(-\frac{1}{2})^n$ and by substitution we have:$$\sum_{n=0}^{\infty}c_nz^n=\sum_{n=0}^{\infty}\frac{2}{3}z^n-\frac{2}{3}(-\frac{1}{2}z)^n$$Convergence of the series requires $|z|<1$ and $|-\frac{1}{2}z|<1$ which finally leads to $|z|<1$ and $R=1$ therefore:$$\sum_{n=0}^{\infty}c_nz^n=\frac{z}{(1-z)(1+\frac{1}{2}z)}\quad,\quad |z|<1$$
