Theorem: Mittag-Leffler's Expansion Theorem

Let $f$ be a meromorphic function that: 
:has only simple poles 
:is continuous, or has a removable singularity, at $0$. 
Let $X$ be the set of poles of $f$. 
For $N \in \N$, let $C_N$ be a disk, centred at the origin, of radius $R_N$ where: 
:$R_N \to \infty$ as $N \to \infty$ 
:$\partial C_N$ contains no poles of $f$ for any $N$. 
:there exists a real number $M > 0$ independent of $N$ such that for all $z \in \partial C_N$, $\cmod {\map f z} < M$ , for all $N \in \N$.
Then:
:$\ds \map f z = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {z - n} + \frac 1 n}$ 
where:
:$\Res f n$ is the residue of $f$ at $n$
:$z$ is not a pole of $f$
:$\ds \map f 0 = \lim_{z \mathop \to 0} \map f z$ if $f$ has a removable singularity at $0$.

Proof:
Let $\zeta \in \C \setminus X$. 
Then: 
:$\ds \frac {\map f z} {z - \zeta}$ 
has simple poles for $z \in X \cup \set \zeta$. 
Let $X_N$ be the set of poles contained within $C_N$. 
Then: 
{{begin-eqn}}
{{eqn | l = \frac 1 {2 \pi i} \oint_{\partial C_N} \frac {\map f z} {z - \zeta} \rd z
      | r = \Res {\frac {\map f z} {z - \zeta} } \zeta + \sum_{n \mathop \in X_N} \Res {\frac {\map f z} {z - \zeta} } n
      | c = Cauchy's Residue Theorem
}}
{{eqn | r = \lim_{z \mathop \to \zeta} \paren {\frac {\paren {z - \zeta} \map f z} {z - \zeta} } + \sum_{n \mathop \in X_N} \paren {\lim_{z \mathop \to n} \frac {\paren {z - n} \map f z} {z - \zeta} }
      | c = Residue at Simple Pole
}}
{{eqn | r = \lim_{z \mathop \to \zeta} \map f z + \sum_{n \mathop \in X_N} \paren {\lim_{z \mathop \to n} \paren {\paren {z - n} \map f z } \cdot \lim_{z \mathop \to n} \frac 1 {z - \zeta} }
      | c = Product Rule for Limits of Real Functions 
}}
{{eqn | r = \map f \zeta + \sum_{n \mathop \in X_N} \frac {\Res f n} {n - \zeta}
      | c = $f$ is continuous at $\zeta$, Residue at Simple Pole
}}
{{end-eqn}}
Setting $\zeta = 0$, we obtain: 
:$\ds \frac 1 {2 \pi i} \oint_{\partial C_N} \frac {\map f z} z \rd z = \map f 0 + \sum_{n \mathop \in X_N} \frac {\Res f n} n$ 
So: 
:$\ds \frac 1 {2 \pi i} \oint_{\partial C_N} \map f z \paren {\frac 1 {z - \zeta} - \frac 1 z} \rd z = \map f \zeta + \sum_{n \mathop \in X_N} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$
It remains to show that the integral on the {{LHS}} vanishes as $N \to \infty$.
We have:
{{begin-eqn}}
{{eqn | l = \cmod {\frac 1 {2 \pi i} \oint_{\partial C_N} \map f z \paren {\frac 1 {z - \zeta} - \frac 1 z} \rd z}
      | r = \cmod {\frac \zeta {2 \pi i} \oint_{\partial C_N} \frac {\map f z} {z \paren {z - \zeta} } \rd z}
}}
{{eqn | o = \le
      | r = \frac {\cmod \zeta} {2 \pi} \cdot \frac M {R_N \paren {R_N - \cmod \zeta} } \cdot 2 \pi R_N
      | c = Estimation Lemma for Contour Integrals and Reverse Triangle Inequality, noting that $\cmod z = R_N$ for $z \in \partial C_N$  
}}
{{eqn | o = \le
      | r = \frac {M \size \zeta} {R_N - \size \zeta}
}}
{{eqn | o = \to
      | r = 0
      | c = as $N \to \infty$, $R_N \to \infty$
}}
{{end-eqn}}
Letting $N \to \infty$ gives: 
:$\ds 0 = \map f \zeta + \sum_{n \mathop \in X} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$
Giving:
:$\ds \map f \zeta = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {\zeta - n} + \frac 1 n}$ 
{{qed}}
{{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}