Radius of convergence of series is $\lim\inf\limits_{n\to\infty}|c_n|^{-1/n}$ 
Let $\{c_n\}$ be a sequence of complex numbers, and $R:=\sup\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}$, which means that R is the radius of convergence of the series $\sum\limits_{n=0}^\infty c_n(z-z_0)^n$ about $z_0$. Show that $R=\lim\inf\limits_{n\to\infty}|c_n|^{-1/n}$.

Unfortunately, I have no idea how to prove this. In fact, the very concept of the limit inferior is very new to me, and this concept is not immediately easy to completely grasp. Moreover, why is $\sup\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}\ne \max_{r\in\mathbb{R}}\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}$, or is it?
I would appreciate some guidance with this problem. In this case I feel completely lost.
 A: 1) $R$ is the radius of convergence:
Proof.
Take $x$ with $0<x<R.$ Then we can find $r$ such that $x<r<R.$ By the definition of $R$, there exists $M$ such that $|c_nr^n|\le M$ holds for all $n$. So we have
$$
|c_nx^n|=|c_n|r^n\left(\frac{x}{r}\right)^n\le M\rho ^n\quad (\rho =x/r<1),
$$
which implies the convergence of $\sum_{n=1}^\infty |c_nx^n|$.
Take $x$  with $x>R$. Then, by the definition of $R$, $\{c_nx^n\}$ is not bounded, so $c_nx^n\to 0$ $(n\to \infty)$ does not hold and we see that the series does not converge.
Thus $R$ is the radius of convergence.
2) $R=\liminf\limits_{n\to\infty}|c_n|^{-1/n}$
EDIT 2
I misunderstood what the author of the book intends. He asks for a direct proof of 
$$
R:=\sup\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}=\liminf_{n\to\infty}\frac{1}{\sqrt[n]{|c_nx^n|}}.
$$
Proof. Take an arbitrary $0\le r <R$. Since $\{c_nr^n\}$ is bounded, there exists $M$ such that 
$$|c_nr^n|\le M.$$
So $$
r \le \left(\frac{M}{|c_n|}  \right)^{\frac{1}{n}}\implies r \le \liminf_{ n\to \infty}\frac{1}{\sqrt[n]{|c_n|}}, 
$$
since $\lim_{ n\to \infty} \sqrt[n]{M}=1.$ Here we used
Theorem. If $r \le a_n\implies r \le \liminf_{ n\to \infty} a_n.$  
Tendeng $r \to R$ we have \begin{align}
R\le \liminf_{n\to\infty}\frac{1}{\sqrt[n]{|c_n|}}.\tag{1}
\end{align}
Next suppose that $$\rho :=\liminf_{n\to\infty}\frac{1}{\sqrt[n]{|c_n|}}>R.$$
Then there exists $N$ such that $$
\frac{1}{\sqrt[n]{|c_n| }}>\frac{\rho +R}{2}$$
holds for all $n>N$. Then we have $$
1>|c_n|\left(\frac{\rho +R}{2}  \right)^n.$$
This is a contradiction. Because $|c_n|\left(\frac{\rho +R}{2}  \right)^n$ is bounded and $\frac{\rho +R}{2}>R$. Thus we can conclude \begin{align}
\liminf_{n\to\infty}\frac{1}{\sqrt[n]{|c_n|}}\le R\tag{2}
\end{align}
and $(1)$ $(2)$ implies 
$$ R=\liminf_{n\to\infty}\frac{1}{\sqrt[n]{|c_n|}}.$$
3) $\sup\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}\ne \max_{r\in\mathbb{R}}\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}$:
Example.
Take $c_n=n.$ Then $\{c_nr^n\}$ is bounded for $r$ with $r<1$ and hence
$$\sup\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}=1.$$
On the contrary,  $\max_{r\in\mathbb{R}}\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}$ does not exist. Indeed 
$$
\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}=[0,1).$$
