# Show that $\lim_{n \to \infty} a_n = \infty$ if and only if $\lim_{n \to \infty} \frac{1}{a_n} = 0$

For a sequence $\{a_n\}$ of positive numbers show that: "$\lim_{n \to \infty} a_n=\infty$ if and only if $\lim_{n \to \infty} \frac{1}{a_n}=0$".

This is the Ex.2.1.17 of Advanced Calculus by P.M.Fitzpatrick's book. As I have learned not to say "Oh, it's trivial" in Mathematics, I tried to prove it but I failed. The problem is that according to the book, "the product of CONVERGENT sequences converges to the product of the limits, and, when ALL quotients are DEFINED, the quotient of CONVERGENT sequences converges to the quotient of the limits". But in this problem, it's not about convergent sequences. Also, The Comparison Lemma doesn't work, in here.

The only approach (for the-left-to-right), I can make is: Suppose there is a sequence of real numbers $\{b_n\}=\{\frac{1}{a_n}\}$; no matter what $a_n$'s are, $\{b_n\}$ is convergent and converges to $0$; then for any $\epsilon>0$, there exist some $N$ such that $|b_n-0|=|\frac{1}{a_n}-0|<\epsilon$ for all indices $n\ge N$. $a_n>0 \implies\frac{1}{a_n}<\epsilon \implies a_n>\frac{1}{\epsilon}$; by the Archimedean Property and since we can choose $\epsilon$ is arbitrary small, so for each positive number $c=\frac{1}{\epsilon}$ there is an index $N$ such that $a_n>c$ for all indices $n\ge N$, then by definition $\lim_{n \to \infty} a_n=\infty$. Is this proof right, especially is it complete with no "gap" (rigorous)?

Here is a better proof. Let $M > 0$ be given, and suppose $\displaystyle \lim_{n\to \infty} \frac{1}{a_n} = 0$ , you must show there exists a positive number $N_0 > 0$ such that : $$n > N_0 \rightarrow a_n > M$$.
Choose $\epsilon = \frac{1}{M}$, then there exists an $N_0$ such that:
$n > N_0 \to \frac{1}{a_n}=\left|\frac{1}{a_n} - 0\right| < \epsilon = \frac{1}{M}\to a_n > M \to \displaystyle \lim_{n\to \infty} a_n = \infty$
$$|a_n| < \varepsilon \Leftrightarrow |\frac{1}{a_n}| > \frac{1}{\varepsilon}$$