= Solution
{c}
All three examples below have positive <integer> <coefficients>, so they work under either convention for whether $0\in\mathbb N$.
For $R=1$, take $\boxed{a_n=1}$. This is the <geometric series>: it converges for $|z|<1$, and its terms fail to tend to zero for $|z|\geq1$.
For $R=0$, take $\boxed{a_n=n!}$, using the <factorial> and $0!=1$. For every fixed $z\ne0$, the <absolute values> of successive terms have ratio $(n+1)|z|$. This eventually exceeds two, so the terms grow without bound and the <series> cannot converge. At $z=0$ it is finite, giving <radius of convergence> zero.
For $R=1/\sqrt2$, take
$$
\boxed{a_n=2^{\lfloor n/2\rfloor},}
$$
using the <floor function>. The <coefficients> satisfy
$$
\frac{(\sqrt2)^n}{\sqrt2}\leq a_n\leq(\sqrt2)^n.
$$
The upper bound and the <comparison test for series> give <absolute convergence> when $\sqrt2|z|<1$. The lower bound prevents the terms tending to zero when $\sqrt2|z|\geq1$. Hence the <radius of convergence> is exactly $1/\sqrt2$.
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