= Solution
{c}
Split the <power series> into even and odd indices:
$$
\sum_{n=0}^{\infty}a_nz^n=\sum_{j=0}^{\infty}\left(\frac z2\right)^{2j}+\sum_{j=0}^{\infty}\frac{z^{2j+1}}{(2j+1)^2},
$$
whenever the two <series> converge absolutely. If $|z|<1$, the even part is a convergent <geometric series>, while the odd part is bounded termwise in <absolute value> by $\sum_j|z|^{2j+1}$, also a convergent <geometric series>. Hence there is <absolute convergence> throughout $|z|<1$.
If $|z|>1$, the odd terms have <absolute values> $|z|^{2j+1}/(2j+1)^2$. Their successive ratio tends to $|z|^2>1$, so these terms grow eventually and do not tend to zero. The <term test for divergence> therefore forces divergence of the original <power series>. Consequently its <radius of convergence> is
$$
\boxed{R=1.}
$$
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