How can I find the radius of convergence of the power series $\sum_{n=1}^{\infty} \frac{n^{n!}}{(n^n)!} x^n$

Question

How can I find the radius of convergence of the following series?

$\sum_{n=1}^{\infty} \frac{n^{n!}}{(n^n)!} x^n$

I was provided the hint: $\forall n \in \mathbb{N}, 3^n (n!) > n^n$

I have no idea about using the hint.

Answer 1

Using the Stirling asymptotic formula

$$ n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n $$

and applying $\log$ to both numerator and denominator ($\log$ is a strictly increasing function) we have

$$ \frac{\log{a_n}}{\log{b_n}} = \frac{1}{e^n}\left(\frac{\sqrt{2\pi n}\log n}{\frac{1/2\log\pi+1/2n\log n}{n^n}+n\log n -1}\right) $$

and clearly

$$ \lim_{n\rightarrow\infty}\frac{\log{a_n}}{\log{b_n}} =0 $$

but

$$ \frac{\log a_n}{\log b_n} = \frac{O(n^{-1/2})}{e^n} $$

then

$$ a_n = b_n^{\frac{O(n^{-1/2})}{e^n}} $$

and then

$$ \frac{a_n}{b_n} = \frac{b_n^{\frac{O(n^{-1/2})}{e^n}}}{b_n}\Rightarrow \lim_{n->\infty}\frac{a_n}{b_n} = 0 $$

because

$$ b_n \approx n \log n-1 $$

hence the convergence radius is $\Re$ or as suggested below, $\infty$

Answer 2

Without Stirling: for large $n$, $n^n - n! > n!$ so we have: $$\frac{n\cdots\hbox{($n!$ times)}\cdots n}{1\cdot 2\cdots (n - 1)n\,\,(n + 1)\cdots(n^n - 1)n^n} < \frac1{n!},$$ and the radius of convergence of $\sum{x^n/n!}$ is infinite, so...