Find the interval of convergence for the series $\sum_{k=0}^\infty a_kx^k$ with $a_k = \alpha a_{k-1} + \beta$ Consider the series
$\sum_{k=0}^\infty a_kx^k$
with $a_0=1, a_k = \alpha a_{k-1} + \beta, k\geq 1, \alpha,\beta \geq 0$.
Determine the interval of convergence of the series.
I've tried looking at the root test and the ratio test, but I don't seem to be getting anywhere. It looks like the expansion for any term $a_n$ is
$a_n = \alpha^n a_{0} + \alpha^{n-1}\beta + \dotsb + \alpha\beta + \beta$, 
but this also seems rather unhelpful.
Any input greatly appreciated!  
 A: You have
$$
a_n = \alpha^n  + \alpha^{n-1}\beta + \dotsb + \alpha\beta + \beta
=\alpha^n+\frac {\beta (1-\alpha^n)}{1-\alpha}
 =\frac {\alpha^n-\alpha^{n+1}+\beta (1-\alpha^n)}{1-\alpha}.
$$ 
For convergence you need $\limsup_n\frac {a_{n+1}x^{n+1}}{a_nx^n}<1$, or $$x <\frac1{\limsup_n \frac {a_{n+1}}{a_n}}=\liminf \frac {a_n}{a_{n+1}}. $$
We have, when $\alpha  <1$,
$$
\frac {a_n}{a_{n+1}}= \frac { \alpha^n-\alpha^{n+1}+\beta (1-\alpha^n)}{ \alpha^{n+1}-\alpha^{n+2}+\beta (1-\alpha^{n+1})}
\xrightarrow [n\to\infty]{} \frac\beta\beta=1.
$$
When $\alpha >1$, 
\begin{align}
\frac {a_n}{a_{n+1}}&= \frac { \alpha^n-\alpha^{n+1}+\beta (1-\alpha^n)}{ \alpha^{n+1}-\alpha^{n+2}+\beta (1-\alpha^{n+1})}
= \frac1\alpha\,\frac { \frac1\alpha-1+\beta (\frac1{\alpha^{n+1}}-\frac1\alpha)}{ \frac1\alpha-1+\beta (\frac1{\alpha^{n+2}}-\frac1\alpha)} \\ \ \\
& \xrightarrow [n\to\infty]{} \frac1\alpha\,\frac { \frac1\alpha-1+\beta (-\frac1\alpha)}{ \frac1\alpha-1+\beta (-\frac1\alpha)} 
=\frac1\alpha.
\end{align}
When $\alpha=1$, we have $a_n =1+n\beta $, so
$$
 \frac {a_n}{a_{n+1}}=\frac {1+n\beta}{1+(n+1)\beta}\xrightarrow [n\to\infty ]{}\frac\beta\beta=1.
$$
In summary, if $\alpha \geq1$ the radius of convergence is $1/\alpha$, while if $\alpha <1$ the radius of convergence is $1$.
