Question regarding big O notation. Let $z\in \mathbb{C}$. I want to show that $\frac{1}{z+n}+\frac{1}{z-n}=O(\frac{1}{n^2})$.
By definition, I need to show that for some constant $M$ and $N$, we have $\mid\frac{1}{z+n}+\frac{1}{z-n}\mid=\mid\frac{2z}{z^2-n^2}\mid\leq \frac{M}{n^2}$ for every $n\geq N$. I need to change this inequality and treat $z$ as a constant. Is this the right approach?

$$\frac1{z+n}+\frac1{z-n}=\frac{z+n+z-n}{(z+n)(z-n)}=\frac{2z}{z^2-n^2}$$
$$\left|\frac{2z}{z^2-n^2}\right|=2|z|\frac{1}{|z^2-n^2|}$$
now we can pick apart this second term:
$$|z^2-n^2|=|n^2[(z/n)^2-1]|=n^2|(z/n)^2-1|$$
so:
$$\left|\frac{2z}{z^2-n^2}\right|=\frac{2}{n^2}\left|\frac{z}{(z/n)^2-1}\right|$$
