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?
On
$$\frac1{z+n}+\frac1{z-n}=\frac{z+n+z-n}{(z+n)(z-n)}=\frac{2z}{z^2-n^2}$$ $$\left|\frac{2z}{z^2-n^2}\right|=2|z|\frac{1}{|z^2-n^2|}$$ 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: $$\left|\frac{2z}{z^2-n^2}\right|=\frac{2}{n^2}\left|\frac{z}{(z/n)^2-1}\right|$$
You have started the right way. Note that $|\frac {2z} {z^{2}-n^{2}}| \leq \frac {2|z|} {n^{2}-|z|^{2}}\leq \frac M {n^{2}}$ if $M( n^{2}-|z|^{2}) \geq 2|z|n^{2}$. This can be written as $n^{2} (M-2|z|) \geq M|z|^{2}$. As long as $M >2|z|$ we can always find $N$ such that this inequality holds for all $n \geq N$. Can you finish?