How to find the radius of convergence of $\sum^{\infty}_{n=1}\frac{n!}{n^n}z^n$?

Question

$$S=\sum^{\infty}_{n=1}\frac{n!}{n^n}z^n$$ Where $z \in \mathbb{C}$. Using D'Alambert's test of convergence:

$$\frac{1}{R}=\lim_{n\to \infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}$$ $$\frac{1}{R}=\lim_{n\to \infty}\frac{n^n}{(n+1)^n}$$

Where $R$ is the radius of convergence. When we consider the difference between $4^3$ and $3^3$, for example, it seems clear to me that this limit cannot be computed as follows:

$$\frac{1}{R}=\lim_{n\to \infty}\frac{n^n}{(n+1)^n}=1$$

However, this is exactly how my textbook (Riley, Hobson and Bence) does it. How can this computation be justified, considering that $n+1$ to a large power will always be a lot larger than $n$.

Answer 1

It is $\lim_{n\to\infty} \frac{n^n}{(n+1)^n}=\lim_{n\to\infty}\left(\frac{n}{1+n}\right)^n=\lim_{n\to\infty}\left(\frac{n+1-1}{n+1}\right)^n=\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)^n=\frac1e$

Answer 2

Hint: $$\frac{n^n}{(1+n)^n}=\frac{1}{(1+\frac{1}{n})^n}$$