Find an $N(\epsilon)$ in this Cauchy sequence proof

Question

This is just a small part of a Cauchy sequence proof I'm doing.

Given an $\epsilon>0$, I need to find an $N(\epsilon)$ so that $$\left|\arctan\left(\frac{n-m}{1+nm}\right)\right|<\epsilon$$ whenever $n,\ m\ge N$.

I get that the denominator of the $\arctan$ argument would be much larger than the numerator but I'm having trouble constructing an $\epsilon-N$ proof.

Answer 1

Hint. Note that if $n\geq m>0$ then $$0\leq \arctan\left(\frac{n-m}{1+nm}\right)\leq \frac{n-m}{1+nm}\leq\frac{1}{m}.$$