<s>[INST] 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. 
 [/INST] 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.
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