The Mittag-Leffler Theorem claims that given a discrete set $S \subset \mathbb{C}$ and principle parts at every $s \in S$, then we can find an analytic function $f: S- \mathbb{C} \to \mathbb{C}$ with $S$ as its set of poles and with the prescribed principle part.
There is a refinement of this theorem which states that we can prescribe for each $s$ not only the principle parts but also finitely many Laurent coefficients for non-negative indices. Under this condition, we can also construct $f$.
I have tried to construct a function which has the form $$ \sum_{s \in S } \left( \sum_{n \geq 1} \dfrac{b_{-n}^{(s)}}{(z-s)^n} - P_s(z) \right) (z-s)^{M_s}$$ where $$(z-s)^{M_s} \sum_{n \geq 1} \dfrac{b_{-n}^{(s)}}{(z-s)^n} $$ is the given restrictions. For example, we can choose for every $s$ a polynomial $P_s(z)$ such that for any $|z| < \dfrac{1}{2} s$, $$\left( \sum_{n \geq 1} \dfrac{b_{-n}^{(s)}}{(z-s)^n} - P_s(z) \right) (2|s|)^{M_s}$$ is sufficiently small, so that the sum converges for all $z$.
I do not find any problems with my proof, but I'm still afraid that it is fake. Also, are there any other approaches to the proof of this theorem?
Say the condition is simple poles of residue $r_n$ at the $a_n,n\ge 1$ where $|a_n| \le |a_{n+1}| \to \infty$
For each $|a_n| > 1$ take $K_n$ such that $r_n/ 2^{K_n} \le 1/n!$ and let $$f(z)=\sum_{n=1}^\infty \frac{r_n}{z-a_n} -P_n(z),\qquad \qquad P_n(z)=\frac{r_n}{-a_n} \sum_{k=0}^{K_n} (z/a_n)^k$$ Can you find if it converges and it is meromorphic ?