Absolute convergence of $\sum_{n\in \mathbb{z}, n\neq 0}^{} \frac{1}{(z-n)n}$ I am tying to show that $\displaystyle \sum_{n\in \mathbb{z}, n\neq 0}^{}   ( \frac{1}{z-n} +\frac{1}{n} )$  is absolutely convergent. I know that equivalently I have to show 
 $\displaystyle \sum_{n\in \mathbb{z}, n\neq 0}^{}  \frac{z}{(z-n)n}$ is absoloutly convrgent. 
I tried to show that
$z \displaystyle \sum_{n\in \mathbb{z}, n\neq 0}^{}  \frac{1}{(z-n)n}$ is convergent and equivalently $\displaystyle \sum_{n\in \mathbb{z}, n\neq 0}^{}  \frac{1}{(z-n)n}$ is convergent. I used this inequality and showed this series is convergent.
$$\sum_{n\in \mathbb{z}, n\neq 0}^{}  \frac{1}{(z-n)n} \leq \sum_{n\in \mathbb{z}, n\neq 0}^{}  \frac{1}{n^2}$$ 
and since the last series is convergent, so the first series is convrgent.
Is there any mistake in this conclusion?
Any help would be great thanks. 
 A: The inequality you displayed isn't exactly accurate, but building on your idea:
$\displaystyle \lim_{n \rightarrow \infty} \frac{(\frac{1}{(z-n)n})}{\frac{1}{n^2}} = -1$
Now, let $\displaystyle \sum_{n=1}^{\infty} \vert g(n)\vert \lt \infty$ be convergent ($\displaystyle g(n) \neq 0$ for any $n$), and suppose $\displaystyle \lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} = M$ where $M \neq 0$. We claim then that $\displaystyle \sum_{n=1}^{\infty} \vert f(n)\vert \lt \infty$.
This can be seen as for $n$ sufficiently large, we must have $\displaystyle \Big\vert\frac{f(n)}{g(n)} - M \Big\vert \lt \epsilon$ for some $\epsilon \gt 0$, so in particular $\displaystyle \vert f(n)\vert \lt (\vert M\vert + \epsilon)\vert g(n)\vert$. Then by comparison we see that $\displaystyle \sum_{n=1}^{\infty} \vert f(n)\vert\lt \infty$.
Putting $\displaystyle f(n) = \frac{1}{(n)(z-n)}$ and $\displaystyle g(n) = \frac{1}{n^2}$, we are in the situation above, and so your series is absolutely convergent for all $\displaystyle z \in \mathbb{C}$.
