limit of $\frac{c^n}{n!}$

Question

Let $c$ be a complex number

Let $u_{n}$ be the sequence defined as $\forall n \in N$ $u_{n}=\frac{c^n}{n!}$

I want to prove that $\lim\limits_{n \rightarrow +\infty} u_{n} = 0$

I would've known to solve this problem if c what replaced by a real number by proving that for $a<n_0<n$;

$0<\frac{a^{n}}{n!}<\ \frac{a^{n-n_{0}}}{(n_{0}-1)!}\times\left(\frac{a}{n_0}\right)^{n-n_{0}+1}$ and then applying the squeeze theorem

By since $c$ is a complex I don't know how to use comparison

I'd like a hint about how to approach it

Thanks a lot for your time

Answer 1

You can apply the same argument in order to prove that $\lim_{n\to\infty}\left|\frac{c^n}{n!}\right|=0$ and$$\lim_{n\to\infty}\left|\frac{c^n}{n!}\right|=0\iff\lim_{n\to\infty}\frac{c^n}{n!}=0.$$

Answer 2

$e^z = \sum_{k=0}^{\infty} \dfrac{z^k}{k!} $is absolutely convergent, $ z \in \mathbb{C}.$

Choose $z=c,$ $ c\in \mathbb {C} :$

$e^c = \sum_{k=0}^{\infty} \dfrac{c^k}{k!}.$

Since absolutely convergent:

$\lim_{k \rightarrow \infty} |\dfrac{c^k}{k!}|=0.$