Show that $\sum_{n = 1}^{\infty} \frac{1}{z + n} + \frac{1}{z-n}$ is absolutely convergent for all $z \in \mathbb{H}$

Question

I need to show that the series $$ \sum_{n = 1}^{\infty} \frac{1}{z + n} + \frac{1}{z-n} $$ is absolutely convergent for all $z$ in the complex upper half plane $\mathbb{H} = \{ z \in \mathbb{C} : \Im(z) > 0 \}$. Taking modulus of each term, I get $$ \sum_{n = 1}^{\infty} \left| \frac{1}{z + n} + \frac{1}{z-n} \right| = 2|z| \sum_{n = 1}^{\infty} \frac{1}{| z^2 - n^2 |}. $$ I think I want to be able to compare this to the series $$ \sum_{n = 1}^{\infty} \frac{1}{n^2} $$ which converges, and so conclude that the original series is absolutely convergent, but I don't know how to do that. Could anyone please help me with this?

Answer 1

Hint. We assume $\Im(z) > 0$. One may write, as $n \to \infty$ $$ \left| \frac{1}{z + n} + \frac{1}{z-n} \right| = \frac1{n^2}\cdot \frac{2|z|}{|\frac{z^2}{n^2}-1 |}\le \frac1{n^2}\cdot \frac{2|z|}{\left|\frac{|z|^2}{n^2}-1\right|} \sim \frac{2|z|}{n^2} $$ giving the absolute convergence for the initial series.