real analysis - If the radius of convergence of $sum^{infty}_{n=1}a_n x^n$ is $R$, what is the radius of convergence of $sum^{infty}_{n=1}a_n x^{n!}$.

If the radius of convergence of $\sum^{\infty}_{n=1}a_n x^n$ is $R$, what is the radius of convergence of $\sum^{\infty}_{n=1}a_n x^{n!}$.

My trial

\begin{align} \limsup\limits_{n\to \infty} \sqrt[n]{|a_n x^{n!}|}&=\limsup\limits_{n\to \infty} \sqrt[n]{|a_n |}\cdot \lim\limits_{n\to \infty} \sqrt[n]{|x|^{n!}}\\&=\dfrac{1}{R}\cdot \lim\limits_{n\to \infty} \sqrt[n]{|x|^{n!}}< 1\end{align}
This implies that

\begin{align} \lim\limits_{n\to \infty} \sqrt[n]{|x|^{n!}}< R&\implies \;{|x|^{(n-1)!}}< R,\;\;\forall\;n\geq N,\;\;\text{for some}\;N,\\&\implies \;{|x|}< R^{1/(n-1)!},\;\;\forall\;n\geq N,\;\;\text{for some}\;N,\\&\implies \;{|x|}\leq 1,\;\;\text{as}\;n\to \infty.\end{align}
I'm I correct? If yes, can I say that the radius of convergence $1?$. If no, can anyone help out or provide a hint?

Answer

Note that $\displaystyle\sum_{n=1}^\infty a_n x^{n!} = \sum_{k=1}^\infty b_k x^k$ where $\displaystyle b_k = \begin{cases} a_n, & k = n!\\ 0, & k \neq n! \end{cases}$

Now you need to consider

$$\limsup_{k \to \infty}\, |b_kx^k|^{1/k} = \limsup_{n \to \infty} |a_n|^{1/n!} |x| < 1,$$

which implies that the radius of convergence is $\left(\limsup_{n \to \infty} |a_n|^{1/n!}\right)^{-1}$.

If $|a_n|^{1/n} \to R^{-1}$ then to what does $|a_n|^{1/n!} = (|a_n|^{1/n})^{1/(n-1)!}$ converge ?