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

$$S=\sum^{\infty}_{n=1}\frac{n!}{n^n}z^n$$ 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 $$n$$.

It is $$\lim_{n\to\infty} \frac{n^n}{(n+1)^n}=\lim_{n\to\infty}\left(\frac{n}{1+n}\right)^n=\lim_{n\to\infty}\left(\frac{n+1-1}{n+1}\right)^n=\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)^n=\frac1e$$
• Where did $e$ come from? Commented May 3, 2019 at 15:24
• $\lim_{n\rightarrow \infty}\frac{n^{n}}{(n+1)^{n}}=\lim_{n\rightarrow \infty}\left ( \frac{n}{n+1} \right )^{n}=\lim_{n\rightarrow \infty}\frac{1}{\left ( 1+\frac{1}{n} \right )^{n}}=\frac{1}{e}$ Commented May 3, 2019 at 15:25
• @Pancake_Senpai It is the definition of $e^x$ as $\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n$. Note, that I had a typo in my calculation, which I edited by now. Commented May 3, 2019 at 15:26
• In this case, it is $x=-1$. So we get $e^{-1}$ as the result. Commented May 3, 2019 at 15:27
Hint: $$\frac{n^n}{(1+n)^n}=\frac{1}{(1+\frac{1}{n})^n}$$