Let \(\omega_1,\omega_2\in\CC\) be linearly independent over \(\RR\), and let \(\Lambda\coloneqq\ZZ\omega_1+\ZZ\omega_2\). The Weierstrass \(\wp\) function of \(\Lambda\) is \[ \wp(z)\coloneqq\frac{1}{z^2}+\sum_{\omega\in\Lambda\setminus\{0\}}\qty{\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}},\qquad z\in\CC\setminus\Lambda. \]
The series defining \(\wp\) converges absolutely and uniformly on compact subsets of \(\CC\setminus\Lambda\), and \(\wp\) is meromorphic on \(\CC\) with a pole of order $2$ at each point of \(\Lambda\) and no other poles.
The number of \(\omega\in\Lambda\) with \(k\le\abs{\omega}<k+1\) is at most \(Ck\) for a constant \(C\) depending on \(\Lambda\), so \(\sum_{\omega\in\Lambda\setminus\{0\}}\abs{\omega}^{-3}<\infty\). Fix \(R>0\). For \(\abs{z}\le R\) and \(\abs{\omega}\ge2R\), \[ \abs{\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}}=\frac{\abs{z}\,\abs{2\omega-z}}{\abs{\omega}^2\abs{z-\omega}^2}\le\frac{R\cdot\frac52\abs{\omega}}{\abs{\omega}^2\cdot\frac14\abs{\omega}^2}=\frac{10R}{\abs{\omega}^3}, \] so by the Weierstrass \(M\)-test the terms with \(\abs{\omega}\ge2R\) converge absolutely and uniformly on \(\abs{z}\le R\) to a holomorphic function. The finitely many remaining terms are holomorphic on \(\abs{z}<R\) away from \(\Lambda\), and the term \((z-\omega)^{-2}\) (or \(z^{-2}\) for \(\omega=0\)) gives a pole of order \(2\) at each \(\omega\in\Lambda\) with \(\abs{\omega}<R\), all other terms being holomorphic near \(\omega\).