Convergence of the sequence $\big(\frac{a^{\underline{n}}}{n!} z^n \big)$ For which values of $z\in\mathbb{C}$ does the sequence $\big(\frac{a^{\underline{n}}}{n!} z^n \big)$ converge for fixed $a\in\mathbb{C}$?
Note here $a^{\underline{n}}$ is the falling factorial i.e. $$a^{\underline{n}}=a(a-1)(a-2)…(a-(n-1)+1)(a-n+1)$$
First observe that if $a$ is a nonnegative integer, $a^{\underline{n}}$ is eventually $0$; thus in that case, the sequence converges for all $z\in\mathbb{C}.$ So assume from now on that $a$ is not a non-negative integer.
The first thing I did was apply the Ratio Test for series which says the radius of convergence of the series $\sum \frac{a^{\underline{n}}}{n!} z^n $ is given by $$R=\lim_{n\to\infty}\frac{\frac{a^{\underline{n-1}}}{(n-1)!}}{\frac{a^{\underline{n}}}{n!}}=1$$ Thus the associated series converges absolutely for $|z|<1\implies$ the sequence $\big(\frac{a^{\underline{n}}}{n!} z^n \big)$ converges to $0$ for all $|z|<1$.
So now it remains to determine if the sequence converges for $|z|\geq1$. It may come in handy later to represent $z$ in polar form: $z=r(\cos\theta + i\sin \theta) \implies z^n=r^n(\cos n\theta +i\sin n\theta)=r^ne^{in\theta}$ where $r\geq1$. Then it remains to compute the limit for $r\geq1:$ $$\lim_{n\to\infty}\frac{a(a-1)(a-2)…(a-(n-1)+1)(a-n+1)}{n!}r^ne^{in\theta}$$
I’m stumped on how to compute this limit other than to possibly convert the $a^{\underline{n}}$ into factorials: $$a^{\underline{n}}=\frac{a!}{(a-n)!}$$ and then possibly use Stirling’s Approximation? I’m not sure what conditions using the factorial formula or Stirling’s Approximation puts on $a$ or $z$, though.
Any hints?
 A: Can anyone verify this estimate?
First we have the identity:
$$\frac{a^{\underline{n}}}{n!}=\frac{a!}{n!(a-n)!}$$ Then by Stirling's Approximation:$$\frac{a!}{n!(a-n)!}\sim\frac{\sqrt{2\pi a}\left(\frac{a}{e}\right)^a}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\cdot\sqrt{2\pi(a-n)}\left(\frac{a-n}{e}\right)^{a-n}}=\frac{1}{\sqrt{2\pi}}\frac{a^{a+1/2}}{n^{n+1/2}(a-n)^{a-n+1/2}}$$
Then, by some questionable algebra I'm hoping folks can check:
\begin{equation} 
\begin{split}
\frac{1}{\sqrt{2\pi}}\frac{a^{a+1/2}}{n^{n+1/2}(a-n)^{a-n+1/2}} & = \frac{a^{a+1/2}}{\sqrt{2\pi}}\frac{1}{n^{n+1/2}\cdot i^{2a-2n+1}(n-a)^{a-n+1/2}} \\
 & = \frac{a^{a+1/2}}{i^{2a-2n+1}\sqrt{2\pi}}\frac{(n-a)^{n-a-1/2}}{n^{n+1/2}}\\
 & = \frac{a^{a+1/2}}{i^{2a-2n+1}\sqrt{2\pi}}\frac{(n-a)^n(n-a)^{-a-1/2}}{n^nn^{1/2}}\\
& = \frac{a^{a+1/2}}{i^{2a-2n+1}\sqrt{2\pi}}\cdot\left(1-\frac{a}{n}\right)^n\cdot\frac{1}{n^{1/2}\cdot(n-a)^{1/2}(n-a)^a}\\
& = \frac{a^{a+1/2}}{i^{2a-2n+1}\sqrt{2\pi}}\cdot\left(1-\frac{a}{n}\right)^n \cdot\frac{1}{\sqrt{n^2-an}\cdot(n-a)^a}
\end{split}
\end{equation}
Thus as $n\to\infty$,
$$\lim_{n\to\infty}\frac{a^{a+1/2}}{i^{2a-2n+1}\sqrt{2\pi}}\cdot\left(1-\frac{a}{n}\right)^n \cdot\frac{1}{\sqrt{n^2-an}\cdot(n-a)^a}$$ $$= \frac{a^{a+1/2}}{\sqrt{2\pi}}\lim\left(\frac{1}{i^{2a-2n+1}}\right)\cdot \lim\left(\left(1-\frac{a}{n}\right)^n\right)\cdot \lim\left(\frac{1}{\sqrt{n^2-an}\cdot(n-a)^a}\right)$$
I can just say the middle limit converges to $e^{-a}$ and the first is periodic and bounded, right?
The last limit is a little spooky. I suppose it would converge to $0$ only if $\Re(a)>-1$, no?
