$\sum_{n=-\infty}^{+\infty}\frac{1}{z-n}$ converges uniformly to a $1-$ periodic meromorphic function. Given a sequence of meromorphic function $(f_n)_n$ on a open set $\Omega\subset \Bbb{C}.$

How does it mean that $\sum_{n=-\infty}^{+\infty}f_n$ converges uniformly on every compact of $D?$

Does it mean that $\sum_{n=1}^\infty (f_n+f_{-n})$ converges uniformly on every compact?

I ask because I find an exercise when it's asking to prove the convergence (on every compact) of $\sum_{n=-\infty}^{+\infty}\frac{1}{z-n}$ to a $1-$periodic meromorphic function. If I guess correctly the definition we have $$\frac1{z-n}+\frac1{z+n}=\frac{2z}{z^2-n^2}.$$

If $A=\sup\{\vert z\vert:z\in K\}$ then $\vert\frac{2z}{z^2-n^2}\vert\le\frac{2A}{n^2-A^2}$ for $n&gt;A.$ So that the serie converges uniformly and it's meromorphic by a theorem.

But why it's $1-periodic?$
$\displaystyle\sum_{n=-\infty}^\infty \frac{1}{z-n}$ diverges but $\displaystyle\frac{1}{2}\sum_{n=-\infty}^\infty (\frac{1}{z-n}+\frac{1}{z+n})$ converges uniformly away from its poles. It is obviously $1$-periodic.
is the sine function remember that $ \frac{1}{z-n}+\frac{1}{z+n} $ is just $ \frac{2z}{z^{2}-n^{2}}$ for every n , now integrate over z and you get

$$ \sum_{n=0}^{\infty}ln(z^{2}-n^{2} $$

which is the logarithm of the sine functio divided by x $ ln(sinx) -ln(x) $