For which complex numbers does a given series converge During studying for an exam, I came across the following exercise:
For which $z \in \mathbb{C}$ does the series
$$\sum\limits_{n=1}^{\infty} \frac{n^n z^n}{n!}$$
converge?
By using the ratio test (or alternatively: calculating the radius of convergence), I can see that the series converges for $|z|<\frac{1}{e}$ and diverges for $|z|>\frac{1}{e}$. However, I am not able to find out what happens if $|z|=\frac{1}{e}$. Does anyone have an idea? 
 A: Claim: if $|z|=\frac1e$ but $z\ne\frac1e$, then the series converges.
Write $z=e^{i\theta-1}$. We can show the claim using Dirichlet's test, with $a_n = n^n/(e^nn!)$ and $b_n = e^{in\theta}$, if we can prove that $\{n^n/(e^nn!)\}$ is a decreasing sequence whose limit equals $0$. (The geometric series formula for $\sum_{n=1}^N e^{in\theta}$ shows that the boundedness condition is satisfied.) Equivalently, one needs to show that
$$
\bigg\{\log \frac{e^nn!}{n^n} \bigg\} = \bigg\{ \sum_{k=1}^n \log k - (n \log n - n) \bigg\} 
$$
is increasing and tends to $\infty$. This can be done by looking at the difference sequence
\begin{multline*}
\bigg( \sum_{k=1}^n \log k - (n \log n - n) \bigg) - \bigg( \sum_{k=1}^{n-1} \log k - ((n-1) \log (n-1) - (n-1)) \bigg) \\
= \log n - \int_{n-1}^n \log x\,dx.
\end{multline*}
That difference is the area of the almost triangle-shaped region above $y=\log x$ and $y=\log n$ between $n-1\le x\le n$; it is certainly greater than the area of the rectangle with upper-left corner $(n-1,\log n)$ and lower-right corner $(n-\frac12,\log(n-\frac12))$, which one can show (using the mean value theorem) is greater than a constant times $\frac1n$. This lower bound for the difference sequence forces the original sequence to increase to infinity by comparison with the harmonic series.
The original series diverges when $z=\frac1e$, but that's difficult to motivate without knowing Stirling's approximation. Nevertheless you can show, using a similar method, that
$$
\frac{n^n}{e^nn!}\cdot n
$$
increases to infinity, which is enough to show that $\sum \frac{n^n}{e^nn!}$ diverges.
A: You need to use something like the Generalized Dirichlet Test (proven in this answer).
This answer shows that $\left(1+\frac1n\right)^n$ is an increasing sequence. Therefore,
$$
\frac{\cfrac{(n+1)^{n+1}}{e^{n+1}(n+1)!}}{\cfrac{n^n}{e^nn!}}=\frac{\left(1+\frac1n\right)^n}e\lt1
$$
and so $\dfrac{n^n}{e^nn!}$ is a decreasing sequence.
When $\left|z\right|=\frac1e$, we can write $z=\frac{e^{i\theta}}e$. Thus,
$$
\begin{align}
\sum_{n=0}^\infty\frac{n^nz^n}{n!}
&=\sum_{n=0}^\infty\frac{n^n}{e^nn!}e^{in\theta}
\end{align}
$$
converges when
$$
\begin{align}
\left|\,\sum_{n=0}^{N-1}e^{in\theta}\,\right|
&=\left|\frac{e^{iN\theta}-1}{e^{i\theta}-1}\right|\\
&\le\frac2{\left|e^{i\theta}-1\right|}\\
&=\frac2{\left|ez-1\right|}
\end{align}
$$
is finite; i.e. $z\ne\frac1e$.
Therefore,
$$
\sum_{n=0}^\infty\frac{n^nz^n}{n!}
$$
converges when $\left|z\right|=\frac1e$, except when $z=\frac1e$.
A: Apply ratio test:
$$\left| {\frac{z \cdot n}{{n+1}}}\right| $$
Converges when $|z|<1$, diverges when $|z|\le1$
