Proof of divergence of two complex series My task is to check, when the following series are convergent:
$$ $$ a) $\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2} z^n$
$$ $$
B) $\sum_{n=1}^{\infty} \frac{n!}{n^n} z^n$
$$ $$ where $z$ is a complex number. Ok, in both cases I computed the radius of convegence. In a) I got $R=\frac{1}{4}$ and in b) I got $R=e$. Those answers are correct, as my book says. But I don't know what happens on the boundary, I mean in first case for $z=\frac{1}{4}$ and in the second for $z=e$. Wolfram says that in both cases it diverges, first one from the comparison test (but compare to what?) and the second one because $\lim \frac{n!}{n^n} e^n = \infty$ (Why?). How to prove it? I really tried, got so frustrated, because it should be easy, but I can't prove divergence of those series... Can you help?

Edit: after hint with Stirling formula, I've got the second one (b) ). Help still needed in a). Thanks in advance.
 A: Based on the comments, you seem to be working under an assumption that if the series has radius of convergence $r$, then its convergence or divergence along the entire boundary can be determined by its convergence or divergence at the point $z=r$. This is false. 
It's easy to see, for example, that for series of the form 
$$\sum_{n=0}^{\infty} a_n z^n$$
with the $a_n>0$ decreasing monotonically from some $n$ on,  and with radius of convergence $0<r<1$,
we must always have convergence at $z=-r$ by the alternating series test.
Don't beat yourself up. When you say "it should be easy" you are wrong. The convergence of a power series along the boundary of its radius of convergence is a complicated question and there are even some open questions remaining for research mathematicians. Are you absolutely sure you have to determine behavior at boundary? 
While this isn't quite an answer, hopefully it should make you feel better!
Edit:
An example often given: The series 
$$
\sum_{n=0}^{\infty} z^n
$$
has radius of convergence $r=1$ but diverges at every point along its boundary, but its derivative
$$
\sum_{n=1}^{\infty} \frac{1}{n} z^n
$$
has the same radius of convergence $r=1$ but diverges at $z=1$ and converges at every other point on the boundary. 
