= Solution
Let $L=\limsup_{n\to\infty}|a_n|^{1/n}$. The <Cauchy-Hadamard theorem>, proved by applying the root test, gives a <radius of convergence>
$$
R=\frac1L
$$
with the usual conventions. For $|z|<R$, choose $q<1$ eventually bounding $|a_nz^n|^{1/n}$, which gives absolute convergence by comparison with a geometric series. For $|z|>R$, infinitely many terms have $n$th root greater than one, so the terms fail to tend to zero and the series diverges.
Here $R=2$ and the assumed limit gives $|a_n|^{1/n}\to1/2$. Therefore
$$
|a_{kn}|^{1/n}
=\left(|a_{kn}|^{1/(kn)}\right)^k
\longrightarrow2^{-k}.
$$
Thus
$$
\boxed{\sum_{n=0}^\infty a_{kn}z^n
\text{ has radius }2^k}.
$$
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