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let $C= \frac{1}{\epsilon}$

There $\exists N\in\mathbb N$ such that for every $n>N$, it is true that: $$a_n>\frac{1}{\epsilon}$$

We should prove that for every $\epsilon>0$ there exists such a $N\in\mathbb N$, for every $n>N$ $$\left|\frac1{a_n}\right|<\epsilon$$

So we take the N that satisfies the first conclusion, and that will mean for every $n>N$$$\left|\frac1{a_n}\right|<\epsilon$$

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  • $\begingroup$ $a_N \gt C$ whenever $n > N$ is probably a typo EDIT: or not, seems quite wayward from the definition of a limit. Go back to the definition of a limit. $\endgroup$
    – Veltas
    Mar 9, 2015 at 17:35
  • $\begingroup$ Careful with the first inequality on your last "therefore" line. Just because $n>N$ doesn't mean that $a_{n}>a_{N}$, i.e. that $\frac{1}{a_{n}}<\frac{a}{a_{N}}$. It is true that if $n>N$ then $a_{n}>C$ and hence $\frac{1}{a_{n}}<\frac{1}{C}$. $\endgroup$
    – TravisJ
    Mar 9, 2015 at 17:41
  • $\begingroup$ I advise you try and make the proof by considering specifically what you need to show: i.e. for an $\epsilon \gt 0$, $\lvert\frac{1}{a_n}\rvert \lt \epsilon$ eventually. $\endgroup$
    – Veltas
    Mar 9, 2015 at 17:46
  • $\begingroup$ "For every $\epsilon \in \mathbb{R}$", probably not the range you want, careful. $\endgroup$
    – Veltas
    Mar 9, 2015 at 17:51
  • $\begingroup$ I think I have it, does it look okay now? @Veltas $\endgroup$ Mar 9, 2015 at 18:09

1 Answer 1

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I would prefer somthing like this (close to what you have):

$\lim\limits_{n\to \infty} a_n=\infty$ means for any $C$ there is an $N$ such that $a_n \gt C$ for $n \gt N$

So given any $\epsilon \gt 0$, let $C = \frac1\epsilon$

and there is an $N$ such that $a_n \gt \frac1\epsilon \gt 0$ for $n \gt N$

i.e. $0 \lt \frac1{a_n} \lt \epsilon$ for $n \gt N$

and that implies $\lim\limits_{n\to \infty} \frac{1}{a_n}=0$

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