$n_{0} \in \mathbb{N}$ such that $\frac{n-m}{mn}<\varepsilon$ for all $n \geq m\geq n_{0}$

Question

I'm trying to prove that a certain sequence is Cauchy and I've managed to, for an arbitrary $\varepsilon>0$, reduce the proof to finding an $n_{0}$ such that, for all $n\geq m \geq n_{0}$, $\frac{n-m}{mn}<\varepsilon$ holds. It seems obvious, but "guessing" such an $n_{0}$ is eluding me.

Answer 1

If $n_0 > \frac{1}{\epsilon}$, then $$ \frac{n-m}{nm} < \frac{n}{nm} = \frac{1}{m} < \frac{1}{n_0} < \epsilon $$

Answer 2

Well, $\frac{n-m}{nm}=\frac1m-\frac1n$. So, take $n_0$ such that $\frac1{n_0}<\varepsilon$. Then, if $n\geqslant m\geqslant n_0$, both $\frac1m$ and $\frac1n$ belong to $\left(0,\varepsilon\right)$ and therefore $\frac1m-\frac1n<\varepsilon$.