= Solution
{c}
For a <power series> $\sum a_nz^n$, its <radius of convergence> $R\in[0,\infty]$ separates <absolute convergence> for $|z|<R$ from divergence for $|z|>R$. One may define $R=\sup\{r\geq0:\sum|a_n|r^n<\infty\}$, allowing $R=\infty$.
To justify convergence inside the radius, choose $|z|<r<R$ with $\sum|a_n|r^n<\infty$. In particular $|a_n|r^n\leq M$ for some $M$, so
$$
|a_nz^n|\leq M(|z|/r)^n.
$$
The <comparison test> with a <geometric series> proves <absolute convergence>. Conversely, convergence at a point of modulus $\rho$ makes $a_nz^n$ bounded there and gives <absolute convergence> at all smaller moduli by the same estimate. Consequently a convergent point of modulus greater than $R$ would contradict the definition of $R$.
For $0<R<\infty$, the real <domain of convergence> is one of $(-R,R)$, $[-R,R)$, $(-R,R]$, or $[-R,R]$. At $R=1$ these are realized respectively by
$$
\sum z^n,\qquad \sum_{n=1}^\infty\frac{z^n}{n},\qquad \sum_{n=1}^\infty\frac{(-1)^nz^n}{n},\qquad \sum_{n=1}^\infty\frac{z^n}{n^2}.
$$
The endpoint assertions follow from the <harmonic series>, <alternating series test>, and <p-series>. Replacing $z$ by $z/R$ realizes every finite positive radius. The degenerate domain $\{0\}$ occurs for $\sum n!z^n$, while the entire <real line> occurs for $\sum z^n/n!$. Their radii are zero and infinity by the <ratio test>.
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