I have been given the above question to prove, I've given it my best shot. Any feedback would be much appreciated.
We are given the following as fact…
"$y_n \rightarrow \infty$ as $n \rightarrow \infty$" means: for each $M \in \mathbb{R}$, there exists $N \in \mathbb{N}$ such that…
$$ n \geq N \implies y_n >M $$
To prove I take the familiar defintion of a limit from the first princibles for all $x_n > 0$ and $\epsilon \in \mathbb{R}^+$.
$$ n \geq N\implies |x_n-0| < \epsilon $$
$$ n \geq N\implies x_n < \epsilon $$
$$ n \geq N\implies \frac{1}{x_n} > \frac{1}{\epsilon} $$
$$ n \geq N\implies \frac{1}{x_n} > M $$
Which matches the above statement we are told is fact, thus proving $\frac{1}{x_n}$ diverges to infinity if its multiplicative inverse converges to zero.
Best Answer
The idea is completely correct. For a neat proof, you should start with: "Let $M>0$." and then pick $\epsilon$ such that $1/\epsilon>M$:
Let $M>0$. There exists $N\in\mathbf{N}$ such that $n\geq N$ implies $x_n<\epsilon$. Since $x_n>0$, we have $\frac{1}{x_n}>\frac{1}{\epsilon}>M$, hence $\frac{1}{x_n}\to \infty$.