Mittag-Leffler on the complex plane

Choose radii $R_n\uparrow \mathrm{\infty }$ so that $A\cap \mathrm{\partial }D(0,R_n)=\mathrm{\varnothing }$ and $A\cap \bar{D}(0,R_n)$ is finite for every $n$. [given, L1, choose] Put $S_1:=A\cap \bar{D}(0,R_1),S_n:=A\cap (\bar{D}(0,R_n)\setminus \bar{D}(0,R_{n-1}))(n\ge 2)\mathrm{.}$ Then each $S_n$ is finite and $A={⨆}_{n\ge 1}{S}_n$. For $n\ge 2$, the finite sum $f_n(z):={\sum }_{a_k\in S_n}{p}_k(z)$ is holomorphic on a neighbourhood of $\bar{D}(0,R_{n-1})$, because every pole it carries lies outside that disc. Set likewise $f_1(z):={\sum }_{a_k\in S_1}{p}_k(z)$.

By [L2], for each $n\ge 2$ choose a polynomial $q_n$ such that [L2, step 1.1, construct] ${\sup }_{\mathrm{\mid }z\mathrm{\mid }\le R_{n-1}}\mathrm{\mid }{f}_n(z)-q_n(z)\mathrm{\mid }<2^{-n}\mathrm{.}$ Put $g_1:=f_1$ and $g_n:=f_n-q_n$ for $n\ge 2$. Then every $g_n$ is meromorphic on $\mathbb{C}$, has the same principal parts as the finitely many $p_k$ with $a_k\in S_n$, and is holomorphic on $\bar{D}(0,R_{n-1})$.

Fix a compact set $K\subseteq \mathbb{C}\setminus A$. Choose $\mathrm{\ell }$ with [step 2.1, choose, algebra, discharge-construct] $K\subseteq \bar{D}(0,R_{\mathrm{\ell }})$. For every $n\ge \mathrm{\ell }+1$, one has $K\subseteq \bar{D}(0,R_{n-1})$, so step 2.1 gives ${\sup }_K\mathrm{\mid }{g}_n\mathrm{\mid }\le 2^{-n}$. Hence ${\sum }_{n\ge 1}{g}_n$ converges uniformly on $K$. Near any point $a_k\in S_{\mathrm{\ell }}$, all terms except $g_{\mathrm{\ell }}$ are holomorphic, so the sum $f:={\sum }_{n\ge 1}{g}_n$ is meromorphic on $\mathbb{C}$ and has principal part $p_k$ at $a_k$.