Is there a nicer way to show that the series is convergent? I'd like to show that for a fixed $z\in\mathbb C\setminus\mathbb Z$ the series $$\sum_{n=1}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right|$$ is convergent.
I think, one can do it as follows. Fix some $n_0> |z|$. Then
\begin{align*}
\sum_{n=1}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| & = \underbrace{\sum_{n=1}^{n_0-1} \left| \frac{1}{z-n} + \frac{1}{n} \right|}_{=:C} + \sum_{n=n_0}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| \\
& = C + \sum_{n=n_0}^\infty \left| \frac{1}{z-n} + \frac{1}{n} \right| \\
& = C + \sum_{n=n_0}^\infty \left| \frac{z}{(z-n)n} \right| \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{|z-n|n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{||z|-|n||n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{(n-|z|)n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{n^2-|z|n} \\
& \leq C + |z|\sum_{n=n_0}^\infty \frac{1}{n^2-|z|n^2} \\
& \leq C + \frac{|z|}{1-|z|} \underbrace{\sum_{n=n_0}^\infty \frac{1}{n^2}}_{<\infty} \\
\end{align*}
That strikes me as somewhat cumbersome. Is there a nicer way? E.g. without separating the series into before and after $n_0$?
 A: We have
$$\left |\frac{1}{z-n} + \frac {1}{n}\right | = \frac{|z|}{n|z-n|}.$$
Divide this by $1/n^2$ to get
$$\frac{|z|}{|z/n - 1|}.$$
The last expression $\to |z|$ as $n\to \infty.$ By the limit comparison test, your series converges (absolutely).
A: Your answer seems to show that the series converges for all $z\in\mathbb{C}$. However, $z$ cannot be a natural number (positive integer).
$$
\begin{align}
\sum_{n=1}^\infty\left|\frac1{z-n}+\frac1n\right|
&=\sum_{n=1}^\infty\left|\frac z{n(z-n)}\right|\\
&=\sum_{n=1}^{\lfloor2|z|\rfloor}\left|\frac z{n(z-n)}\right|
+\sum_{n=\lfloor2|z|\rfloor+1}\left|\frac z{n(z-n)}\right|\\
&=\underbrace{\sum_{n=1}^{\lfloor2|z|\rfloor}\left|\frac z{n(z-n)}\right|}_{\text{finite for $z\not\in\mathbb{N}$}}
+\left|2z\right|\underbrace{\sum_{n=\lfloor2|z|\rfloor+1}^\infty\frac1{n^2}}_{\le\frac{\pi^2}6}
\end{align}
$$
A: I don't know about elegance, but you can also rewrite $z$ as $x+iy$ to get
$\sum_{n=1}^\infty \sqrt{[\frac{x-n}{(x-n)^2+y^2}+\frac{1}{n}]^2+O(\frac{1}{n^4})} $
Now $\frac{x-n}{(x-n)^2+y^2}+\frac{1}{n}$ can be simplified as $\frac{x^2-xn+y^2}{n[(x-n)^2+y^2]}=O(\frac{1}{n^2})$ and the rest is simple.
