Think about what the Ratio Test says. Given a power series
$$\sum_{n=0}^\infty a_n x^n$$
the ratio test says that the series converges absolutely if
$$L = \lim_{n\to\infty} \left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| |x| < 1$$
(it may or may not converge absolutely if that is $1$, but this won't affect the radius of convergence)
Equivalently, we have that
$$|x| < \frac{1}{\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|}$$
And since $R$ is defined to be the radius of the largest disk (interval) in which the series converges absolutely, this tells us that
$$R = \frac{1}{\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|}$$
and if this limit exists and is non-zero, this tells us in particular that the radius of convergence $R$ is finite, and in this case, is given by precisely
$$R = \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|$$
Concerning the example, try $$\sum_{n=0}^\infty n!\cdot x^n$$
In this case, $a_n = n!$, and $R = \lim_{n\to\infty} \left|\frac{n!}{(n+1)!}\right| = 0$, but $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ does not exist, but this yields $R = 0$, and the series converges only at $0$.
If instead, you want a series where the limit $\lim_{n\to\infty} \left|\frac{a_n}{a_{n+1}}\right|$ doesn't exist, take
$$\sum_{n=0}^\infty \frac{x^n}{n!}$$
i.e., $a_n = \frac{1}{n!}$, and in this case, $$\lim_{n\to\infty} \left|\frac{a_n}{a_{n+1}}\right| = \lim_{n\to\infty} \left|\frac{(n+1)!}{n!}\right| = \lim_{n\to\infty} (n+1)$$
which fails to exist, but does in fact yield $R = \infty$, since $\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = 0$ is the reciprocal of the radius of convergence.
It's not true that if $|z|\limsup|c_{n+1}/c_n|>1$ then the series diverges.
In fact there are big problems here if $c_n=0$ for some $n$. But that's not the only problem. Consider the sequence
$$(c_n)=(1,2,1/4,2/4,1/9,2/9\dots),$$
or $c_{2k}=1/(k+1)^2$, $c_{2k+1}=2/(k+1)^2$. Let $z=1$. Then $|z|\limsup|c_{n+1}/c_n|=2$, but $\sum c_nz^n$ converges.
Best Answer
The answer to question 1. is yes.
Let $R=|z-a|$. Since $\lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|=\infty$ there is some $N$ such that $\left|\frac{c_n}{c_{n+1}}\right| > 2R$ for all $n\geq N$. By induction $\left|\frac{c_N}{c_n}\right| > (2R)^{n-N}$.
Thus $$\sum_{n=N}^\infty |c_n (z-a)^n| = |c_N|\sum_{n=N}^\infty\left|\frac{c_n}{c_N}\right|R^{n} < |c_N|\sum_{n=N}^\infty (2R)^{N-n}R^{n} $$ $$= |c_N|R^{N}\sum_{n=0}^\infty 2^{-n} = 2|c_N|R^N$$ so the series converges absolutely.
For a concrete counterexample for 2. take $c_n = \begin{cases} 1 & n\mbox{ even} \\ \frac{1}{n} & n\mbox{odd}\end{cases}$ and note that the series clearly diverges for $|z-a|\geq1$.