1.3 Theorem. For a given power series $\sum_{n=0}^\infty a_n(z-a)^n$ define the number $R$, $0 \le R \le \infty$, by $$\frac{1}{R} = \limsup |a_n|^{1/n}.$$ Then...(b) if $|z-a| > R$, the terms of the series become unbounded and so the series diverges...
I left out parts (a) and (c) as they didn't give me very much trouble. WLOG, take $a=0$. Here is Conway's proof of part (b):
Here is my attempt at filling in the details:
Recall that if $a < \limsup x_n$, then for every $N \in \Bbb{N}$, there is $n \ge N$ s.t. $x_n > a$. Hence, given $N=1$, there is $n_1 \ge 1$ s.t. $|a_{n_1}| > \frac{1}{r^{n_1}}$. If $n_1=1$, take $N=2$. Then there is $n_2 \ge 2 > n_1$ s.t. $|a_{n_2}| > \frac{1}{r^{n_2}}$. However, if $n_1 > 1$, then take $N = n_1 - 1$. Either way, continuing this process gives us a strictly increasing sequence $\{n_k\} \subseteq \Bbb{N}$ for which $|a_{n_k}| > \displaystyle \frac{1}{r^{n_k}}$ for every $k \in \Bbb{N}$. Then $|a_{n_k}z^{n_k}| > \left(\frac{|z|}{r}\right)^{n_k}$, where $\frac{|z|}{r} > 1$. This implies
$$\sum_{n=0}^\infty |a_{n}z^{n}| \ge \sum_{k=0}^\infty |a_{n_k}z^{n_k}| \ge \sum_{k=0}^\infty \left(\frac{|z|}{r}\right)^{n_k} = \infty,$$
but why does this imply $\sum_{n=0}^\infty a_{n}z^{n}$ diverges?
The subsequence $(a_{n_k}z^{n_k})$ is unbounded, hence $(a_nz^n)$ does not converge to $0$, thus $\sum_{n=0}^\infty a_{n}z^{n}$ is divergent.