I'm trying to prove that the radius of convergence of $\sum \frac{n!}{n^n} a_n z^n$ is $eR$ when $R>0$ is the radius of convergence of $\sum a_n z^n$.
I can easily obtain that the radius of convergence of $\sum \frac{n!}{n^n} a_n z^n$ is $\ge eR$ thanks to a more general rule. But how can I manage to get the equality (without using Stirling formula if possible)?
Using Striling Formula, one can prove that since $\frac{n!}{n^n} a_n \sim \frac{a_n}{e^n}\sqrt{2\pi n}z^n$, $\sum \frac{n!}{n^n} a_n z^n$ has same radius than $\sum \frac{a_n}{e^n}\sqrt{2\pi n} z^n$ which has same radius than $\sum \frac{a_n}{e^n}z^n$ which is $eR$.
Note that $$\lim _{n\to \infty }{a_n\over a_{n+1}}=R$$therefore $$\lim_{n\to \infty}{{n!\over n^n}a_n\over {(n+1)!\over (n+1)^{n+1}}a_{n+1}}{=\lim_{n\to \infty } {1\over n+1}{ (n+1)^{n+1}\over n^n}\lim_{n\to \infty }{a_n\over a_{n+1}}\\=\lim_{n\to \infty } { (n+1)^{n}\over n^n}\lim_{n\to \infty }{a_n\over a_{n+1}}\\=eR}$$