Write $R(a)$ for the radius of convergence of the series $\sum\limits_na_nx^n$. Introduce $b_n=(a_n)^{\color{green}{\alpha}}$ for a given positive integer $\color{green}{\alpha}$ and write $R(b)$ for the radius of convergence of the series $\sum\limits_nb_nx^n$.
Lemma: Assume there exists a number $\varrho$ such that: (1) if $|x|\lt\varrho$, then $a_nx^n$ converges to $0$, and (2) if $|x|\gt\varrho$, then $a_nx^n$ does not converge to $0$. Then $R(a)=\varrho$.
Application:
- Assume that $|x|\lt R(a)^\color{green}{\alpha}$. Then $|x|^{1/\color{green}{\alpha}}\lt R(a)$, hence $a_n|x|^{n/\color{green}{\alpha}}$ converges to $0$. Thus $b_nx^n$ converges to $0$ since $|b_nx^n|$ is the $\color{green}{\alpha}$th power of the modulus of $a_n|x|^{n/\color{green}{\alpha}}$.
- Assume that $|x|\gt R(a)^\color{green}{\alpha}$. Then $|x|^{1/\color{green}{\alpha}}\gt R(a)$, hence $a_n|x|^{n/\color{green}{\alpha}}$ does not converge to $0$. Thus $b_nx^n$ does not converge to $0$ since $|b_nx^n|$ is the $\color{green}{\alpha}$th power of the modulus of $a_n|x|^{n/\color{green}{\alpha}}$.
Thus: $\color{red}{R(b)=R(a)^\color{green}{\alpha}}.$
Proof of the lemma: Assume that (1) and (2) hold.
- Let $x$ such that $|x|\lt\varrho$. There exists $\sigma$ such that $|x|\lt\sigma\lt\varrho$. By (1), $a_n\sigma^n$ converges to $0$, in particular $|a_n\sigma^n|\leqslant C$ for every $n$. Thus $|a_nx^n|\leqslant Cu^n$ with $u=|x|/\sigma\lt1$, hence $\sum\limits_na_nx^n$ converges. Since this holds for every $|x|\lt\varrho$, $R(a)\geqslant\varrho$.
- Let $x$ such that $|x|\gt\varrho$. By (2), $a_nx^n$ does not converge to $0$, hence $\sum\limits_na_nx^n$ diverges, which means that $R(a)\leqslant|x|$. Since this holds for every $|x|\gt\varrho$, $R(a)\leqslant\varrho$. QED.