Proof explanation: $\lim\limits_{n\to\infty}\frac{z^n}{n!}=0 $ for $z\in \mathbb{C}$

Question

$\lim\limits_{n\to\infty}\frac{z^n}{n!}=0 $ for $z\in \mathbb{C}$

Proof:

Let $m:=\lfloor|z|\rfloor $, then for $n>m$:$$\left \lvert \frac{z^n}{n!}\right \rvert=\frac{|z|^m}{m!}\cdot \frac{|z|^{n-m}}{(m+1)\cdot \ldots \cdot n}\leq |z|^m\cdot \left \lvert \frac{z}{m+1}\right \rvert ^{n-m}$$

Since $ \left \lvert \frac{z}{m+1}\right \rvert<1$ and $|z|^m$ being a constant, we can conclude that as $n\to \infty$ we get $0$.

I don't understand why $m$ is defined like that... why does it matter in the proof?

Answer 1

$m$ is defined the way it is because that's the exact point where the increase in $n!$ becomes larger than the increase in $|z^n|$ as $n$ grows, so it's at that point the absolute value of the fraction starts shrinking. You could use a larger $m$, if you wanted. But for a smaller $m$, the fraction would still be increasing in size.

Answer 2

Using $m$ as the modulus of $z$ is not the only way to prove the limit, but it's likely one of, if not the, easiest ways. The reason they define it like this is to show that, no matter how large $\left|z\right|$ is, eventually $n!$ will become infinitely larger than $\left|z^n\right|$. This is through the manipulation into an inequality with $2$ factors, one which is a constant and the other going to $0$ in modulus as $n \to \infty$.

Answer 3

Another way that highlights the same idea...

Pick a positive integer $m$ such that $m > |z|$. Then, for $n>m$, \begin{multline*} \ln(n!\,/\,\left|z\right|^{n})=\ln(n!)-n\ln\left|z\right|=\left(\ln n+\cdots+\ln m\right)+\left(\ln(m-1)+\cdots+\ln2\right)-n\ln\left|z\right|\\ \geq\left(n-m\right)\ln m-n\ln\left|z\right|=n\left(\ln m-\ln\left|z\right|\right)-m\ln m=nC_{1}-C_{2} \end{multline*} where $C_{1}>0$ and $C_{2}$ are constants depending on $|z|$. Exponentiating, we see that $n!\,/\,|z|^{n}$ diverges.

Moral of the story. Sometimes it's easier to think about these things from the additive perspective.