I have $$ \overline{\Bbb C^2}\ni(x,y)\mapsto\cfrac{\epsilon_{2n+1}}{(x-a_{2n+1})+\cfrac{\epsilon_{2n}}{(y-a_{2n})+\cfrac{\epsilon_{2n-1}}{\dots}}} $$ where $\overline{\Bbb C^2}$ is the one-point compactification of $\Bbb C^2$, that is $\overline{\Bbb C^2}=\Bbb C^2\cup\{\infty\}$, where the symbol $\infty$ denotes any couple of complex numbers $(x,y)$ such that at least one of them is the point at infinity of the Riemann Sphere.
Call the above map $f_{2n+1}$, where $\{a_n\}_n$ is a sequence of complex numbers, such that $|a_n|$ strictly increases and diverges.
I want to study convergence of such a sequence of functions, to a function $$ f:\overline{\Bbb C^2}\to\overline{\Bbb C}\;. $$
Intuitively, taking the $\epsilon_j$ going to $0$ fast enough should guarantee convergence.
Looking here it seems that $|x-a_{2n+1}|\ge \epsilon_{2n+1}+1$ and $|y-a_{2n}|\ge \epsilon_{2n}+1$ should guarantee convergence, but it's not really clear (also, it talks about real numbers).
Moreover since I am considering mappings $\overline{\Bbb C^2}\to\overline{\Bbb C}$ (in particular the value $\infty\in\overline{\Bbb C}$ is allowed, is a point like others), we can state I am interested in convergence to a non-constant function (in particular $f\not\equiv\infty$). Moreover, the sequence $a=\{a_n\}_n$ is divergent; all these pieces of information led me to wonder if the choice of $\epsilon_j$ can be made independently from $a$.
Or, the other way around: maybe the convergence to a constant mapping is a particular case, hence one asks wether it does exist a sequence $\{\epsilon_n\}_n$ such that the sequence $\{f_n\}_n$ doesn't converge/converge to a constant mapping.
Can someone give some hint/reference/solution? Thanks