TITLE: Explanation needed for a statement about power series convergence
QUESTION [5 upvotes]: I got a task in front of me but I don't really understand it. If someone could explain, I think I would be able to solve it myself.

$P(x) = \sum_{k=0}^{\infty}a_{k}x^{k}$ is a power series. There exists a $k_{0} \in \mathbb{N}$ with $a_{k} \neq 0$ for all $k \geq k_{0}$.
Proof that: If the sequence $\left ( \left | \frac{a_{k+1}}{a_{k}} \right | \right )_{k \geq k_{0}}$ converges towards a number in $\mathbb{R}$ or towards $\infty$ and if $a:= \lim_{k\rightarrow \infty} \left | \frac{a_{k+1}}{a_{k}} \right | \in \mathbb{R} \cup \left \{ \infty \right \}$ indicates this limit point ($\infty $ or $-\infty$) then following applies for the radius of convergence $R$ of $P$:
$R=\left\{\begin{matrix}
0, & a = \infty\\ 
\infty, & a = 0 \\ 
\frac{1}{a}, & otherwise 
\end{matrix}\right.$


What is meant by $k_{0}$ ? It's just any unknown variable which seems to be smaller or equal $k$, right? Oh and it cannot be smaller than zero.
What is $a_{k}$ ? It's just any sequence that cannot be zero, right?
So first I take the sequence $a_{n}$, use the ratio test to see if it converges. Okay after that is done, I check if in the ratio test, I get + or - $\infty$.
Is it right so far?
But what confuses me most is this: 

$a:= \lim_{k\rightarrow \infty} \left | \frac{a_{k+1}}{a_{k}} \right | \in \mathbb{R} \cup \left \{ \infty \right \}$

What is it saying with infinity?
Sorry I haven't started with the task but first I try to understand everything, then start.

REPLY [3 votes]: Hint: Since you've gotten explanation of the notation already, here's the sketch of how to prove the result. Radius of convergence means that the power series converges for anything strictly inside. So if $|x| = r < \frac1a$, then let $s$ be such that $r < s < \frac1a$, and so as $k \to \infty$ eventually $|\frac{a_{k+1}}{a_k}| \to a$ and hence $|\frac{a_{k+1}}{a_k}| < s^{-1}$, which by induction gives $|a_{m+k}| < |a_m| s^{-k}$ for every natural $k$ where $m$ is some (sufficiently large) constant natural number. Therefore for any natural $q \ge p \ge m$ we have $| \sum_{k=p}^q a_k x^k | \le \sum_{k=p}^\infty |a_k| r^k = \sum_{k=p}^\infty |a_m| s^{m-k} r^k = |a_m| s^m \sum_{k=p}^\infty |a_m| (\frac{r}{s})^k$ which is finite since $\frac{r}{s} < 1$. Thus by Cauchy convergence the original power series converges.