Prove that if $a_n$ tends to $\infty$ then $\frac{1}{a_n}$ tends to 0

Question

I have the definition that a sequence to tends to $\infty$ if, for every $C>0$, there exists a natural number $N$ such that $a_n > C$ for all $n>N$

And I have the definition that a sequence tends to $0$ if, for each $\epsilon >0$, there exists a natural number $N$ such that $|{a_n}|< \epsilon$ for all $n>N$

I don't know quite how to form a concise yet rigorous proof.

Answer 1

Let $ \epsilon >0$ and put $C=1/ \epsilon$. There is $N$ such that $a_n >C$ for $n>N$, hence $1/a_n < \epsilon $ for $n>N$.