Is there a nicer way to show that the series is convergent?

Question

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 $n_0$?

Answer 1

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).

Answer 2

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.

Answer 3

Your answer seems to show that the series converges for all $z\in\mathbb{C}$. However, $z$ cannot be a natural number (positive integer). $$ \begin{align} \sum_{n=1}^\infty\left|\frac1{z-n}+\frac1n\right| &=\sum_{n=1}^\infty\left|\frac z{n(z-n)}\right|\\ &=\sum_{n=1}^{\lfloor2|z|\rfloor}\left|\frac z{n(z-n)}\right| +\sum_{n=\lfloor2|z|\rfloor+1}\left|\frac z{n(z-n)}\right|\\ &=\underbrace{\sum_{n=1}^{\lfloor2|z|\rfloor}\left|\frac z{n(z-n)}\right|}_{\text{finite for $z\not\in\mathbb{N}$}} +\left|2z\right|\underbrace{\sum_{n=\lfloor2|z|\rfloor+1}^\infty\frac1{n^2}}_{\le\frac{\pi^2}6} \end{align} $$