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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 "$n > \text{something}$".

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    $\begingroup$ How about ${1 \over n!} \le {1 \over n}$. $\endgroup$
    – copper.hat
    Commented Nov 18, 2015 at 7:44
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    $\begingroup$ The absolute value doesn't matter here since $n!>0$ for all $n \in \mathbb{N}$. $\endgroup$
    – MrMazgari
    Commented Nov 18, 2015 at 8:25
  • $\begingroup$ 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 $n! \geq \frac{1}{\epsilon}$. $\endgroup$
    – EHH
    Commented Nov 18, 2015 at 9:24

1 Answer 1

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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.

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  • $\begingroup$ I think you want $\varepsilon<1$. In either case, I don't see how it adds or takes away any simplicity. $\endgroup$
    – Wojowu
    Commented Jan 18, 2016 at 16:30
  • $\begingroup$ @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. $\endgroup$
    – Snow
    Commented Jan 18, 2016 at 16:40
  • $\begingroup$ 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. $\endgroup$
    – Wojowu
    Commented Jan 18, 2016 at 16:45

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