If $\displaystyle a_n$ > 0 $\displaystyle \forall$ n $\displaystyle \in$ $\displaystyle \mathbb{N}$ and $\displaystyle \lim_{n\rightarrow\infty}$ $\displaystyle a_n$ = 0, prove that $\displaystyle \dfrac{1}{a_n}$ $\displaystyle \rightarrow$ $\displaystyle \infty$.

I haven't been able to write up a proof for this one yet. I have a few ideas down:

Since {$\displaystyle a_n$} $\displaystyle \rightarrow$ 0, by definition of convergence, I know |$\displaystyle a_n$| < $\displaystyle \varepsilon$, or

-$\displaystyle \varepsilon$<$\displaystyle a_n$<$\displaystyle \varepsilon$.

Starting ideas for proof:

Assume $\displaystyle \lim_{n\rightarrow\infty}$ $\displaystyle a_n$ = 0 and $\displaystyle a_n$ > 0 $\displaystyle \forall$ n $\displaystyle \in$ $\displaystyle \mathbb{N}$.

{$\displaystyle a_n$} $\displaystyle \rightarrow$ 0 $\displaystyle \Rightarrow$|$\displaystyle a_n$ - 0| < $\displaystyle \varepsilon$ $\displaystyle \Rightarrow$ |$\displaystyle a_n$| < $\displaystyle \varepsilon$ $\displaystyle \Rightarrow$ |$\displaystyle \dfrac{1}{a_n}$| > $\displaystyle \dfrac{1}{\varepsilon}$, since $\displaystyle \dfrac{1}{\varepsilon}$ > 0 .....

I was working with my classmates, and we were guessing that we would reach a contradiction somehow using the definition of convergence, and then by ______ (we don't know what exatly) we will conclude that the sequence has to diverge by contradiction to our assumption.

Any help, suggestions, tips, corrections, guidance, etc. is greatly appreciated. Thank you very much for looking and for your time!