sequence $a_n =\sum_{k=0}^{n-1}(-1)^k\frac{(n-k)^k}{k!}$ converge or diverge?

Question

How check that sequence $$a_n =\sum_{k=0}^{n-1}(-1)^k\frac{(n-k)^k}{k!}$$ is converge or diverge ?

Answer 1

Since, $$\displaystyle \begin{align} f(z) = \dfrac{1}{1-ze^{-z}} = \sum_{k=0}^{\infty} z^k.e^{-kz} &= \sum_{k=0}^{\infty} z^k.\left(\sum_{j=0}^{\infty} (-1)^{j} \dfrac{k^{j}}{j!}z^j\right) \\&= \sum_{k=0}^{\infty}\sum_{j=0}^{\infty} (-1)^j \dfrac{k^j}{j!}z^{k+j}\end{align}$$

Changing the summation with $n = j+k$:

$$ = \sum_{n=0}^{\infty} \left(\sum_{j=0}^{n-1}(-1)^{k}\dfrac{(n-j)^j}{j!}\right)z^n = \sum_{n=0}^{\infty} a_n z^n$$

Then, $\displaystyle \sum_{n=0}^{\infty} a_n$ converges and in turn $a_n \to 0$, if $f(z)$ is analytic in a disc centered at origin and of radius greater than $1$.

So, it suffices to prove that every singularity of $f(z)$ appears outside the unit disc $|z| > 1$.

Suppose, there is a singularity of $f(z)$ inside the disc $|z| \le 1$, say at $z = z_0 = x_0 + iy_0$,

then, $z_0 = e^{z_0} \iff x_0 + iy_0 = e^{x_0 + iy_0} = e^{x_0}(\cos y_0 + i\sin y_0)$ where, $x_0^2 + y_0^2 \le 1$

i.e., $x_0 = e^{x_0}\cos y_0 \textrm{ and } y_0 = e^{x_0}\sin y_0 \implies x_0 = \frac{y_0}{\tan y_0} > 0$ (since, $|y_0| \le 1 < \frac{\pi}{2}$)

Therefore, $1 \ge |z_0| = e^{x_0}|e^{iy_0}| = e^{x_0} > 1$, contradiction.

Hence, $f(z)$ is analytic in a disc $|z| > 1$.

Therefore, $\displaystyle \lim\limits_{n \to \infty} \sum_{k=0}^{n-1}(-1)^k\frac{(n-k)^k}{k!} = 0$.