= Solution
{c}
Use the <ceiling function> to choose positive <integer> <coefficients>
$$
\boxed{a_n=\lceil\alpha^{-n}\rceil\qquad(n\geq0).}
$$
Because $0<\alpha<1$, we have $\alpha^{-n}\geq1$ and
$$
\alpha^{-n}\leq a_n<\alpha^{-n}+1\leq2\alpha^{-n}.
$$
For $|z|<\alpha$, the <comparison test for series> with the <geometric series> gives
$$
\sum_{n=0}^{\infty}|a_nz^n|\leq2\sum_{n=0}^{\infty}\left(\frac{|z|}{\alpha}\right)^n<\infty.
$$
For $|z|\geq\alpha$, we instead have $|a_nz^n|\geq(|z|/\alpha)^n\geq1$, so the terms do not tend to zero. Thus \b[the <power series> has <radius of convergence> exactly $\alpha$.]
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