Convergence of the series $\sum_{n=0}^\infty \frac{n^n z^n}{n!}$ when $e|z|=1$ Using the ratio test or any other methods, one can easily find the radius of convergence of the following series to be $1/e$.
$$
\sum_{n=0}^\infty \frac{n^n z^n}{n!}
$$
When $z=1/e$, wolfram tells me it diverges, although I cannot prove why it diverges. For $z=-1/e$, it is an alternating series, so it definitely converges. It should also be convergent when $z=\pm i/e, i^2=-1$.
Let the set $S$ be the set of angles $\theta\in (-\pi,\pi]$ for which the series converges when $z=e^{i\theta-1}$. I want to find the set $S$. I am guessing that $S=\{\theta\in(-\pi,\pi]:\theta\neq 0\}$, but I don't know how to prove it.
Can this be generalised to more general series?
 A: As is quite frequently the case on the boundary of convergence, this is a job for Dirichlet's test:

If 
  
  
*
  
*$(a_n)$ is a real sequence with $a_n \downarrow 0 $
  
*$(b_n)$ is a complex sequence with bounded partial sums, $ \left\lvert \sum_{k=0}^n b_k   \right\rvert < M$ for some fixed $M$,
  
  
  then $ \sum_{n} a_n b_n $ converges.

Namely, since
$$ \frac{n^n e^{-n}}{n!} \downarrow 0 , $$
(the ratio of successive terms is $ (1+1/n)^n/e < 1$, see below for a better estimate) and
$$ \left\lvert \sum_{k=0}^n e^{in\theta} \right\rvert = \left\lvert \frac{1-e^{i(n+1)\theta}}{1-e^{i\theta}} \right\rvert < \csc{(\theta/2)} $$
is bounded for $\theta \neq 0$, we do indeed have convergence for all points on the circle apart from $\theta=0$.
As for divergence for $\theta=0$, there are various ways to do this, but probably the simplest is to derive a lower bound for $n^n e^{-n}/n!$ and apply the comparison test. The graph of $\log{x}$ is concave, so the Trapezium Rule gives an underestimate of its area; this can be massaged to give the following lower bound:
$$ \frac{1}{2}\log{1} + \log{n!} - \frac{1}{2}\log{n} < \int_1^n \log{x} \, dx = 1-n + n\log{n} , $$
so
$$ n! < e\sqrt{n} n^n e^{-n} . $$
Therefore
$$ \frac{n^n e^{-n}}{n!} > \frac{1}{e\sqrt{n}} , $$
and finally $\sum_n n^{-1/2}$ diverges by comparison with the harmonic series, for example.
