= Solution
For either exponent, the recurrence from part (c) implies
$$
\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right|
=\frac{|x|}{|[2(n+1+\sigma)-1](n+\sigma)|}
\longrightarrow0
$$
for every finite complex $x$. The <ratio test> therefore gives
$$
\boxed{R=\infty\quad\text{for both power-series factors}}.
$$
The factor multiplying $x^\sigma$ is an <entire function> in each case. For $\sigma=1/2$, this does not make the full <Frobenius solution> single-valued at zero: its square-root prefactor still requires a branch. The infinite radius refers to $\sum a_nx^n$, not to a Taylor expansion of the branched full solution.
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