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

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?

• Try to estimate $|n^2-z^2|\ge n^2-|z|^2$, so for $n>|z|$ your series terms are majorated by $\frac{1}{n^2-|z|^2}$.
– A.Γ.
Nov 7 '16 at 17:44

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.
• Why do you need $\Im(z) > 0$?
• @A.G. If $\Im(z) = 0$ then there is a possibility that the sum is not well-defined, as the denominator can become zero for some $n$. Otherwise, I don't think it plays any role.
• @Brahadeesh It does not matter that the denominator is not defined for finitely many $n$. What matters is that it is defined for all $n$ large enough. It is pointwise absolute convergence question.
• I just gave the same frame: the OP question is to prove it for $z \in \mathbb{H}$. Nov 7 '16 at 19:06