Radius of convergence for $\sum_{n=1}^\infty \frac{n^n}{n!}z^n$ Find the radius of convergence for: $$\sum_{n=1}^{\infty} \frac{n^{n}}{n!}z^{n} $$
My attempt: 
Let $$c_{n}=\frac{n^{n}}{n!}$$
$$\limsup_{n \to \infty} |c_{n}|^{\frac{1}{n}}=\limsup_{n \to \infty} \frac{n}{(n!)^{1/n}}$$
$$=\limsup_{n \to \infty} \frac{n}{n^{1/n}{(n-1)^{1/n}(n-2)^{1/n}\cdots 2^{1/n}1^{1/n}}}$$
The denominator $$n^{1/n}{(n-1)^{1/n}(n-2)^{1/n}\cdots 2^{1/n}1^{1/n}}$$ goes to 1 and so the limit goes to $\infty$
So the radius of convergence is $1/\infty=0$
Is that right? 
 A: Using ratio test would be easy here. Consider
$$\frac{c_n}{c_{n+1}} = \frac{n^n}{n!}\frac{(n+1)!}{(n+1)^{n+1}} = \frac 1 {(1+\frac{1}{n})^n}$$
$$\limsup_{n \rightarrow \infty} \left| \frac{c_n}{c_{n+1}} \right| = \frac 1 e$$
So the radius of convergence is $\frac{1}{e}$.
A: By Lagrange inversion theorem, the Lambert function $W$ has the following Taylor series at the origin:
$$ W(x) = \sum_{n\geq 1}\frac{(-1)^{n+1}n^{n-2}}{(n-1)!}\,x^n $$
whose radius of convergence equals $\frac{1}{e}$. It leads to 
$$ \frac{W(x)}{1+W(x)} = \sum_{n\geq 1}\frac{(-1)^{n+1}n^{n}}{n!}\,x^n$$
then to:
$$ \sum_{n\geq 1}\frac{n^n}{n!}\,x^n = \frac{-W(-x)}{1+W(-x)} $$
for any $|x|<\frac{1}{e}$.
A: No, this is not correct. Applying the ratio test,
$$
\frac{(n+1)^{n+1}n!}{(n+1)!n^n}|z|=\frac{(n+1)(n+1)^n}{(n+1)n^n}|z|=\left (\frac{n+1}{n}\right )^n|z|=\left (1+\frac{1}{n} \right )^n|z|.
$$
Since the first term tends to $e$ for $n \to \infty$, the ratio is strictly smaller than one iff $|z|<e^{-1}$.
Your mistake comes from the fact that it is true that individual terms at the denominator go to zero, but the number of individual terms grows with $n$, so you cannot apply that reasoning.
A: This is more easily done with the ratio test.
$\frac {(n+1)^{n+1}} {(n+1)!} \frac {n!} {n^n} = (\frac {n+1} {n})^n$
which goes to $e$ as $n$ goes to infinity, so the radius of convergence is $\frac {1} {e}.$
A: 
If one wishes to use the root test, then one can proceed as follows. 

Note that
$$\begin{align}
\lim_{n\to \infty}\frac n{(n!)^{1/n}}&=\lim_{n\to \infty}\left(\prod_{k=1}^{n-1}\left(1-\frac kn\right)\right)^{-1/n}\\\\
&=\lim_{n\to \infty}e^{-\frac1n \sum_{k=1}^n\log\left(1-\frac kn\right)}\\\\
&=e^{-\lim_{n\to \infty}\frac1n \sum_{k=1}^n\log\left(1-\frac kn\right)}\\\\
&=e^{\int_0^1 \log(1-x)\,dx}\\\\
&=e^{-1}
\end{align}$$
Hence, the series converges for $|z|<1/e$.
