radius of convergence for $\sum_{n=1}^{\infty} \frac{z^{n} n^{n}}{n!}$ and $\sum_{n=1}^{\infty} z^{n!}$ Exercise 4:10 in John D'Angelo's text is to find the radius of convergence for :
A) $\sum_{n=1}^\infty \frac{z^n n^n}{n!}$ and 
B)  $\sum_{n=1}^\infty z^{n!}$
I got half of an answer for A) which I wanted to check and I got totally stuck on B). Thanks for the help.
I know from the Theorem in the section that $\frac{1}{R} = \limsup |a_n|^{\frac{1}{n}}$ where $R$ is the radius of convergence. So, 
for A:
$$\frac{1}{R} = \frac{n}{n!^{\frac{1}{n}}}\text{ so }\frac{1}{R} = \limsup \frac{n!^\left({\frac{1}{n}}\right)}{n}$$ which I think is $0$ but I'm not positive. I doubt this is correct because that would mean that the radius of convergence is $\infty$ which seems wrong.
for B: 
$z^{n!} = z^{(n \times(n-1)!)}$ but I don't know how to eliminate the $(n-1)!$ so I
can just have a $z^{n}$ so that I can use the theorem above regarding $R$.
Thanks again. Oh, I know I asked this before but if anyone knows of a solution manual to this text, I'd appreciate it. I'm not a student so not trying to cheat on the homework but rather just trying to understand the basics.
 A: For A), try the ratio test instead: ${1 \over R} = \lim_n  | {a_{n+1} \over a_n } | = \lim_n { (n+1)^{n+1} \over (n+1)!} { n! \over n^n } = \lim_n (1+ {1 \over n})^n = e$.
For B), it might be easier to compute $\limsup_n \sqrt[n]{|z|^{n!}} = \limsup_n |z|^{(n-1)!}$. For $|z|<1$ this is zero, for $|z|>1$ this is infinite.
A: Alternative proof for (B): as
$$\sum_{n=1}^N |z|^{n!}\le\sum_{n=1}^{N!} |z|^n,$$
the radius of convergence is $\ge 1$. But
$$\sum_{n=1}^\infty 1^{n!}$$
diverges, so the radius is $=1$.
A: Regarding problem A:
Put $$a_n = \frac{n^n}{n!}$$
By the Cauchy D'Alembert criterion we have
$$\frac{1}{R}=\lim_{n\to\infty}\sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n}=\lim_{n\to\infty}\frac{(n+1)^n}{n^n}=\lim_{n\to\infty}\Big(1+\frac{1}{n}\Big)^n=e\\\iff R=\frac{1}{e}.$$
Regarding problem B:
Put $$b_n=z^{n!}.$$
Then $$\sqrt[n]{b_n}=|z|^{n!/n}$$ which converges to $\infty$ if $|z|>1$ and to $0$ if $|z|<1$. Thus the radius of convergence is $1$.
