Given is a sequence $(z_n)_n$ of distinct complex numbers with $|z_n|\to +\infty$ as $n\to +\infty$ and a sequence of singular parts $$ B_n(z):=\frac{b_{1,n}}{z-z_n}+\dots+\frac{b_{N_n,n}}{(z-z_n)^{N_n}},$$ then there exists a meromorphic function on $\mathbb{C}$ for which $z_n$ are the only poles and $B_n(z)$ the singular part of the Laurent series in $z_n$, for all $n\in\mathbb{N}$.
Proof:
$B_n(z)$ is holomorphic on $\mathbb{C}\backslash\{z_n\}$ and therefore its Taylor series around $0$ converges in $B(0,|z_n|)$. Moreover, it converges uniformly in $B(0,|z_n|/2)$. $\color{red}{\text{This implies that there exists a Taylor polynomial } P_n \text{ of } B_n \text{ around 0}}^{[1],[2]}$ for which $$ |B_n(z)-P_n(z)|\le \frac1{2^n},\forall z\in B\left(0,\frac{|z_n|}2\right).$$ Now, we have to show that $f(z):=\sum_{n=1}^{\infty} (B_n(z)-P_n(z))$ converges. Choose $M\in\mathbb{N}$, then $$ \sum_{n:|z_n|\ge 2M}(B_n(z)-P_n(z))$$ $\color{red}{\text{converges uniformly on } B(0,M)}^{[3]}$. Therefore, the sum is holomorphic on $B(0,M)$. We obtain $f$ by adding the remaining terms (with $|z_n|<2M$, finite number since $|z_n|\to\infty$). Consequently, $f$ is holomorphic on $B(0,M)\color{red}{\backslash\{z_n:n\in\mathbb{N}\}}^{[4]}$.
Choose $n_0\in\mathbb{N}$, then $f(z)=B_{n_0}(z)-P_{n_0}(z)-\sum_{n\ne n_0}(B_n(z)-P_n(z))$, so that the singular part of the Laurent-series of $f$ in $z_{n_0}$ is only given by $B_{n_0}(z)$.
[1]: How does uniformly convergence imply this existence?
[2]: What should we do in the case of $z_n=0$. How do we handle $B(0,|z_n|)$ and Taylor-series around this pole?
[3]: I can see how this has to relate to the M-test (with $\sum 1/2^n$), but I'm not sure how to explain that the convergence disc is $B(0,M)$.
[4]: Why is this set excluded? In the previous sentence, we had established that the sum was holomorphic on the whole disc $B(0,M)$, so why are we now avoiding the $z_n$'s?
Thanks.
Best Answer
Definition of uniform convergence, with $\epsilon = 1/2^n$.
There's nothing special about $0$. If some $z_n = 0$, translate:
consider instead $z'_n = z_n - d$ for some $d$ that is not one of the $z_n$, then translate back at the end.
Who says the convergence disk is $B(0,M)$? It converges there, but also elsewhere.
Because $B_n(z)$ has a singularity at $z=z_n$. In the previous sentence, you had $f$ which didn't have that singularity, now you have added it in.