# Prove that if $\lim_{n\to \infty} a_n=\infty$ then $\lim_{n\to \infty} \frac{1}{a_n}=0$

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$$\left|\frac1{a_n}\right|<\epsilon$$ •$a_N \gt C$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. Mar 9, 2015 at 17:35 • 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$\frac{1}{a_{n}}<\frac{1}{C}$. Mar 9, 2015 at 17:41 • 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. Mar 9, 2015 at 17:46 • "For every$\epsilon \in \mathbb{R}\$", probably not the range you want, careful. Mar 9, 2015 at 17:51
• I think I have it, does it look okay now? @Veltas Mar 9, 2015 at 18:09

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