Suppose $g(u)=\sum_{i=1}^\infty f(i) \frac{1}{u-f(i)}$
How do I obtain $f$ given $g$?
Example pair: $g(u)=\frac{\sqrt{u}-\pi \cot \left(\frac{\pi }{\sqrt{u}}\right)}{2 \sqrt{u}}, f(i)=i^{-2}$
This is a Stieltjes transformation which has an inverse formula in wikipedia but we can't take the limit here. Is there an approach that works for the example pair?
Notice that it is impossible to exactly get $f$ from $g$ because $g$ deosn't change when we permute the terms of $f$. Indeed, for any permutation $\sigma$ of $\mathbb{N}^*$ and any $u$, $\sum_{i \geqslant 1} \frac{f(i)}{u - f(i)} = \sum_{i \geqslant 1} \frac{f(\sigma(i))}{u - f(\sigma(i))}$ assuming that we have absolute convergence. Therefore, the best we result we could obtain is to know for each $u \in \mathbb{C}$ the number of $i \in \mathbb{N}^*$ such that $f(i) = u$. Let us call $c(u)$ this number. Notice also that adding some $0$ in the sequence $f$ doesn't change $g$ so we may assume without loss of generality that $f$ doesn't vanish (hence $c(0) = 0$).
Assume that $g$ exists for at least one $u_0 \in \mathbb{C}$. Then, for all $i \in I$, $u_0 \neq f(i)$ and $\frac{f(i)}{u_0 - f(i)} = \frac{1}{\frac{u_0}{f(i)} - 1} \rightarrow 0$ when $i \rightarrow +\infty$. Therefore, $\left|\frac{u_0}{f(i)}\right| \rightarrow +\infty$ so $u_0 \neq 0$ and $f(i) \rightarrow 0$. It implies by the way that $c(z) < +\infty$ for any $z \neq 0$. I will assume for simplicity that the previous sequence converges absolutely, whcih means that $\sum_{i \in I} \left|\frac{f(i)}{u_0 - f(i)}\right| < +\infty$. As $f(i) \rightarrow 0$, $\left|\frac{f(i)}{u_0 - f(i)}\right| \sim \frac{|f(i)|}{|u_0|}$ hence $f \in \mathcal{l}^1$.
Let $\varepsilon > 0$ and $i_\varepsilon$ be some index such that $\forall i \geqslant i_\varepsilon, |f(i)| < |u_\varepsilon|$. Let $A_\varepsilon = \{u \in \mathbb{C}||u| \geqslant \varepsilon, \forall i \leqslant i_\varepsilon - 1, |u - f(i)| \geqslant \varepsilon\}$. Then, for all $u \in A_\varepsilon$, \begin{align*} \sum_{i \geqslant 1} \left|\frac{f(i)}{u - f(i)}\right| & = \sum_{i = 1}^{i_\varepsilon - 1} \frac{|f(i)|}{|u - f(i)|} + \sum_{i \geqslant i_\varepsilon} \frac{1}{\left|\frac{u}{f(i)} - 1\right|}\\ & \leqslant \sum_{i = 1}^{i_\varepsilon - 1} \frac{|f(i)|}{\varepsilon} + \sum_{i \geqslant i_\varepsilon} \frac{1}{\left|\frac{u}{f(i)}\right| - 1}\\ & \leqslant \frac{1}{\varepsilon}\sum_{i = 1}^{i_\varepsilon - 1} |f(i)| + \sum_{i \geqslant i_\varepsilon} \frac{1}{\left|\frac{\varepsilon}{f(i)}\right| - 1}. \end{align*} This sum doesn't depend on $u$ and since $f(i) \rightarrow 0$, $\frac{1}{\frac{\varepsilon}{|f(i)|} - 1} \sim \frac{|f(i)|}{\varepsilon}$ so the series converges. It proves that $g$ converges normally on each $A_\varepsilon$. Therefore, it exists on the open set $U = \bigcup_{\varepsilon > 0} A_\varepsilon = \mathbb{C}\backslash\{f(\mathbb{N}^*) \cup \{0\}\}$ and the convergence is locally uniform. In particular, $g$ is holomorphic (it is a consequence of Morera theorem for example). If $u \in \mathbb{C}^*$, we can use the same method to prove that $z \mapsto (z - u)g(z)$ is holomorphic arround $u$ because the problematic terms $\frac{f(i)}{z - f(i)}$ where $f(i) = u$ in the sum are cancelled by the factor $z - u$. We deduce that $g$ is meromorphic on $\mathbb{C}^*$ with only simple poles.
Let $u \in \mathbb{C}^*$. Then, as $f(\mathbb{N}^*)\backslash\{0\}$ is discrete in $\mathbb{C}^*$ (because $f$ converges toward $0$), there exists a $\varepsilon > 0$ such that $f(\mathbb{N}^*) \cap \overline{B}(u,\varepsilon) \subset \{u\}$. Let $\gamma_u : t \mapsto u + \varepsilon e^{2\mathbf{i}\pi t}$, defined for $0 \leqslant t \leqslant 1$. We have $\mathrm{Ind}_u(\gamma_u) = 1$ and for all $z \in f(\mathbb{N}^*) \cup \{0\}$, $\mathrm{Ind}_z(\gamma_u) = 0$. Moreover, the function $u \mapsto \sum_{i \geqslant 1} \left|\frac{f(i)}{u - f(i)}\right|$ is continuous on $U$ by local uniform convergence thus, $$ \int_{\gamma_u} \sum_{i \geqslant 1} \left|\frac{f(i)}{z - f(i)}\right| \, dz < +\infty. $$ We deduce by Fubini theorem that, $$ \int_{\gamma_u} g(z) \, dz = \sum_{i \geqslant 1} \int_{\gamma_u} \frac{f(i)}{z - f(i)} \, dz = \sum_{i \geqslant 1} 2\mathbf{i}\pi f(i)\delta_{f(i),u} = 2\mathbf{i}\pi uc(u). $$ And, $$ \int_{\gamma_u} g(z) \, dz = 2\mathbf{i}\pi\mathrm{Res}_u(g), $$ by the residue theorem. We deduce that for all $u \neq 0$, $$ \fbox{$\#(f^{-1}(\{u\})) = \frac{\mathrm{Res}_u(g)}{u}$} $$