Radius of convergence of $\sum\limits_n a_n^3x^n$ Given a series $\sum\limits_{n=0}^{\infty}a_nx^n$ with radius of convergence $R$, I need to find the radius of convergence of the series $\sum\limits_{n=0}^{\infty}a_n^3x^n$.
I would like to use d'Alembert theorem and say: $R'=\lim _{n \to \infty}\left|\frac{a_n^3}{a_{n+1}^3}\right|=R^3$. What's wrong with that? What is the correct answer?
Thank you!!
 A: 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.

A: If you have the formula $R = \liminf_{n \rightarrow \infty} |a_n|^{-{1 \over n}}$, then the new radius of convergence is $\liminf_{n \rightarrow \infty} |a_n^3|^{-{1 \over n}} = 
\liminf_{n \rightarrow \infty} (|a_n|^{-{1 \over n}})^3 = (\liminf_{n \rightarrow \infty} |a_n|^{-{1 \over n}})^3 = R^3$. 
But to use the above you should show that the liminf of the cubes of a sequence of nonnegative real numbers is the cube of the liminf. 
