# Finish proving that $\lim_{n\to \infty}\frac 1{n!}=0$

Prove that: $$\lim\limits_{ n \to \infty}\frac{1}{n!}=0.$$

I tried this:

$$\left\vert \frac{1}{n!}-0\right\vert < \varepsilon$$ $$\frac{1}{n!} < \varepsilon$$ $$\varepsilon * n! > 1$$ $$n! > \frac{1}{\varepsilon} .$$ Now what can I do in order to get an expression like this: $n > \text{something}$, without the absolute value? According to my teacher we always have to get "$n > \text{something}$".

• How about ${1 \over n!} \le {1 \over n}$. Nov 18, 2015 at 7:44
• The absolute value doesn't matter here since $n!>0$ for all $n \in \mathbb{N}$. Nov 18, 2015 at 8:25
• I assume it's what copper.hat is getting at but basically since $n! \geq n$, then if we choose $n$ s.t. $n\geq \frac{1}{\epsilon}$ then we also have $n! \geq \frac{1}{\epsilon}$.
– EHH
Nov 18, 2015 at 9:24

To complete the proof, we say: Choose $n> \dfrac 1\varepsilon$. Now, if $n>\dfrac 1\varepsilon$ then $$\frac 1{n!}<\frac 1n <\frac 1{\frac{1}{\varepsilon}}=\varepsilon.$$ This completes the proof.
• I think you want $\varepsilon<1$. In either case, I don't see how it adds or takes away any simplicity. Jan 18, 2016 at 16:30
• @Wojowu, why do you want $\varepsilon < 1$? If it is larger, then $\frac 1\varepsilon < 1$ and you can use $n=1$. If you want to be pedantic, you can say that you want the existence of a natural number $N$ so that the inequality is true for all $n>N$. This follows from the Archimedean Property.
• The content of definition of limit is that no matter how small $\varepsilon$ is, then we have $N$ such that... you know the rest. Hence assuming $\varepsilon$ is big (i.e. bigger than $1$) we can't fulfill that definition. Jan 18, 2016 at 16:45