Proof of an inequality about $\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2}$ I've encountered an inequality pertaining to the following expression:
$\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2}$, where $z$ is a complex number.
After writing $z$ as $x + iy$ we have the inequality when $y \gt 1$ and $|x| \le \frac{1}{2} $:  
$|\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2}|
 \le C + C\sum_{n=1}^{\infty}\frac{y}{y^2+n^2}$  
The "proof" of the inequality is given as follows:
  $\frac{1}{z} + \sum_{n=1}^{\infty}\frac{2z}{z^2 - n^2} = \frac{1}{x+iy} +\sum_{n=1}^{\infty}\frac{2(x+iy)}{x^2 - y^2 - n^2 + 2ixy}$  
But I fail to see how the inequality follows. 
 A: As
$$1\leq y\leq|x+iy|\leq{y\over2}+y={3\over2} y$$
we have
$${1\over |x+iy|}\leq 1$$
and
$$|n^2+y^2-x^2-2ixy|\geq |n^2+y^2-x^2|\geq\Bigl(1-{1\over8}\Bigr)(n^2+y^2)\qquad(n\geq1)\ .$$
It follows that
$$\left|{1\over z}+\sum_{n=1}^\infty{2z\over z^2-n^2}\right|\leq 1+\sum_{n=1}^\infty {3y \over{7\over8}(n^2+y^2)}=1+{24\over7}\sum_{n=1}^\infty {y \over n^2+y^2}\ .$$
Therefore the stated inequality is true with $C=4$, say.
A: The following is not the simplest way to solve this question but it may provide additional insight.
The key observation is that both sums have a closed-form representation. To see this, we first need to show that for $w = \sigma + it$
$$ |\pi \cot(\pi w)| \le \pi \coth(\pi t).$$
This is because
$$ |\pi \cot(\pi w)| = 
\pi \left| 
\frac{e^{i\pi\sigma-\pi t}+e^{-i\pi\sigma+\pi t}} 
{e^{i\pi\sigma-\pi t}-e^{-i\pi\sigma+\pi t}}
\right|\le \pi
\frac{e^{\pi t}+e^{-\pi t}}{e^{\pi t}-e^{-\pi t}} = \pi \coth(\pi t)$$
for $t>0.$
Similarly, when $t<0$,
$$ |\pi \cot(\pi w)| = 
\pi \left| 
\frac{e^{i\pi\sigma-\pi t}+e^{-i\pi\sigma+\pi t}} 
{e^{i\pi\sigma-\pi t}-e^{-i\pi\sigma+\pi t}}
\right|\le \pi
\frac{e^{\pi t}+e^{-\pi t}}{e^{-\pi t}-e^{\pi t}} = -\pi \coth(\pi t)$$
Now we use a classic technique to evaluate the two sums, introducing the functions
$$ f_1(w) = \pi \cot(\pi w) \frac{2z}{z^2-w^2} \quad \text{and} \quad
f_2(w) = \pi \cot(\pi w) \frac{z}{z^2+w^2},$$
with the conditions that $z$ not be an integer for $f_1(z)$ and not $i$ times an integer for $f_2(z).$ We choose $\pi \cot(\pi w)$ because it has poles at the integers with residue $1$.
The key operation of this technique is to compute the integrals of $f_1(z)$ and $f_2(z)$ along a circle of radius $R$ in the complex plane, where $R$ goes to infinity. Now by the first inequality we certainly have $|\pi\cot(\pi w)| < 2\pi$ for $R$ large enough. The two terms $\frac{2z}{z^2-w^2}$ and $\frac{z}{z^2+w^2}$ are both $\theta(1/R^2)$ so that the integrals are $\theta(1/R)$ and vanish in the limit. (Here we have used the two bounds on $|\pi \cot(\pi z)|$ that we saw earlier.) This means that the sum of the residues at the poles add up to zero.
Now let
$$ S_1 = \sum_{n=1}^\infty \frac{2z}{z^2-n^2} \quad \text{and} \quad
S_2 = \sum_{n=1}^\infty \frac{y}{y^2+n^2}.$$
By the Cauchy Residue Theorem,
$$ \frac{2}{z} - 2\pi\cot(\pi z) + 2S_1 = 0  \quad \text{and} \quad
 \frac{1}{y} - \pi\coth(\pi y) + 2S_2 = 0.$$
Solving these, we obtain
$$ S_1 = -\frac{1}{z}  + \pi\cot(\pi z)   \quad \text{and} \quad
 S_2 = - \frac{1}{2} \frac{1}{y} + \frac{1}{2} \pi\coth(\pi y).$$
Starting with the left side of the original inequality we finally have 
$$\left| \frac{1}{z} + S_1\right| < \pi \coth(\pi y) = 2 S_2 +  \frac{1}{y} .$$
It follows that $C=2$ is an admissible choice.
