From the wiki page, it looks like a meromorphic the function is written as $$f(z) = \sum_{a\in E}p_a(z)$$ where $p_a(z)$ is the principal part (a Laurent series), $a$ is the place where a singularity occurs and I is the points belonging to some open set where the singularities occur. It looks like in the theorem, it also defines $p_a(z) = \sum_{n=1}^{N_a}\frac{c_{a,n}}{(z-a)^n}$
So let's say I have one of the examples on the page $\cot(z)$, we would have a singularity at the points $z=0$ and $z=n\pi$ and this would occur for all integers $-\infty \le n \le \infty$, so if I were to write this out, I would get:
$$f(z) = \cot(z) = \frac{\cos(z)}{\sin(z)} = \lim_{N\rightarrow\infty} \sum_{n=-N}^N \Big( \sum_{n=1}^{N_a}\frac{c_{n\pi,n}}{(z-n\pi)^n}\Big)$$
I'm assuming that after finding $p_a(z)$ from finding the Laurent series (however that's done, I've never done one myself), the sum ends up becoming: $$f(z) = \lim_{N\rightarrow \infty} \sum_{n=-N}^N \frac{1}{z-n\pi}$$.
Is that correct?
Note: Something unusual I noticed in the example for $\cot(z)$ is that despite $\frac{1}{z-n\pi}$ not being even (nor odd), it appears that the infinite sum was simplified by being split and doubled i.e.: $$\sum_{-N}^N a_n = a_0 + 2\sum_{n=1}^N a_n$$.
Best Answer
The Mittag-Leffler theorem (as per Shabat's "Introduction to complex analysis") states the following:
Theorem (Mittag-Leffler)
In plain language, it says that for every possible sequence of poles and principal parts there exists a meromorphic function that has exactly these poles and principal parts. Note that this function is not unique: since entire functions (i.e. holomorphic in $\mathbb{C}$) have no poles and principal parts in $\mathbb{C}$, if $f(z)$ is a function the existence of which is stated by Mittag-Leffler theorem, then for any entire $h(z)$ $f(z)+h(z)$ has the same poles and principal parts, and thus verifies the theorem. Mittag-Leffler theorem has a corollary which allows one to express an arbitrary meromorphic function in terms of its principal parts.
Corollary
Several remarks about this corollary.
If a function $f(z)$ has only a finite number of poles, $f(z)-\sum_{n}g_{n}(z)$ is well-defined and is an entire function, so in the case of a meromorphic function with a finite number of poles there is no need for correction polynomials. Hence, for a meromorphic function $f(z)$ with a finite number of poles, $$f(z)=h(z)+\sum_{n=1}^{N}g_{n}(z)$$ where $h(z)$ is an entire function, and $g_{n}(z)$ are principal parts of $f(z)$. In particular, any rational function can be decomposed into a polynomial (the entire part) and principal parts (partial fraction decomposition).
In the case of an infinite number of poles we need the polynomials $P_n(z)$, because otherwise the series may not converge. These polynomials, as was mentioned before, are, in general, obtained in the following way: for each principal part $g_{n}(z)$ at the point $z_{n}$, we decompose it in the Taylor series $$\sum_{k=0}^{\infty} \frac{d^n}{dz^n}g_{n}(z)\Bigg|_{z=0} \cdot \frac{z^k}{k!}$$ around zero. We can do that, because $g_{n}(z)$ has a single pole at $z_n$, hence, the radius of convergence of its Taylor series around zero is $|z_n|$. We then truncate the Taylor series for $g_{n}(z)$ around such term that inside the circle $|z|\le |z_{n}|$ $\left|g_{n}(z)-\sum_{k=0}^{N(n)}\frac{g^{(k)}(0)}{k!}z^{k} \right|$ is sufficiently small, for instance, less than $\frac{1}{2^{n}}$. This is possible because the Taylor series of a function holomorphic at a certain domain converges uniformly inside this domain, and $g_{n}(z)$ is holomorphic in the disk $|z|<|z_{n}|$. Then the series $\sum_{n=0}^{\infty}(g_{n}-P_{n}(z))$ converges uniformly on every compact in $\mathbb{C}\backslash \{z_{n}\}_{n=1}^{\infty}$. The choice of truncation number is not unique thanks to the freedom of choice of the entire function $h(z)$.
In the proof of the Mittag-Leffler theorem we use a system of expanding contours around zero, the interior of which in the limit covers the whole complex plane and which have a "well-behaved" boundary (means it does not become a fractal in the limit). The series of principal parts, in fact, must be summed in the same order the corresponding poles were covered by the contours, because the series does not have to converge absolutely. In the explanation above that system was just the sequence of circles $|z|=|z_n|$, but we may choose a different system, if we make the series converge uniformly inside each contour. If $f(z)$ behaves on these contours in a nice way, that allows us to simplify the Mittag-Leffler series.
For instance, if $f(z)=o(1)$ at a certain system of such expanding contours (decresases at infinity), then, by Liouville's theorem, the entire part of $f(z)$ is zero (because all entire functions are either constants or grow in absolute value at infinity, and the maximum of modulus is always reached at the boundary of each domain), so the Mittag-Leffler series becomes the sum of principal parts: $$f(z)=\sum_{n=0}^{\infty}g_{n}(z)$$ Again note that the principal parts must be summed in order they appear inside an expanding sequence of contours.
If $f(z)$ grows as a polynomial $|z^p|$ on a certain system of contours, then we can represent $$f(z)=f^{[p-1]}(z)+\sum_{n=0}^{\infty}(g_{n}(z)-g_{n}^{[p-1]}(z))$$ where the notation $f(z)^{[k]}(z)$ denotes the Taylor series for $f(z)$ at zero, truncated at the $k$-th term (the term with $z^{k}$ is included, the next one is not).
As an example, consider the function $\frac{1}{\sin(z)}$. It has a pole at zero, $1/\sin(z) \approx 1/z$ around zero, so, for the reasoning behind the theorem to work, we have to first subtract $\frac{1}{z}$, which is the principal part at 0. For the new function $\frac{1}{\sin(z)}-\frac{1}{z}$, which has poles at the points $\pm \pi k$, $k\in \mathbb{N}$, we can choose the sequence of square contours $C_{n}$ which are squares with the side length $2\pi n + 1$ centered at zero and passing through the real line vertically at points $\pm \pi(n + \frac{1}{2})$. At these points $\frac{1}{\sin(z)}-\frac{1}{z}$ has values $1+\frac{\pm 1}{\pi k}$, and outside the real axis the absolute value decreases as $1/|z|$, since $1/|\sin(z)|$ decays exponentially outside the real axis. So our function is $o(z^{1})$ at the chosen contours, and we can write down decomposition: $$\frac{1}{\sin z}-\frac{1}{z}=\underbrace{\left(\frac{1}{\sin z}-\frac{1}{z}\right)(0)}_{\text{The 1-1=0-th term of the Taylor decomposition}} = 0$$ $$ + \lim_{N\to \infty} \sum_{n=1}^{N} \left(\text{Res}_{z=\pi n} \left(\frac{1}{\sin(z)} -\frac{1}{z}\right) \frac{1}{z-\pi n} + \text{Res}_{z= -\pi n}\left(\frac{1}{\sin(z)} - \frac{1}{z} \right) \frac{1}{z+\pi n}\right)$$
where we used the fact that zeros of entire function $f(z)$ are exactly the poles of $\frac{1}{f(z)}$ of the same degree (or just the $2\pi k$-translational invariance of $\sin(z)$) to guess the placement and degree of the poles. Note also that the polynomials corresponding to principal parts are of degree 0, hence, they are the zeroth terms of Taylor expansion of $\frac{1}{1 \pm \pi k z}$ around zero, and cancel out for each pair $\pm k$. Now, let us calculate the residues: $$\text{Res}_{z=\pm \pi n} \frac{1}{\sin(z)}-\frac{1}{z}=\lim_{z\to \pm \pi n}(z- \mp \pi n)\left( \frac{1}{\sin(z)} - \frac{1}{z} \right)=(-1)^{n}$$ The fact that the values of the residue alternate can be seen from L'Hopital's rule and the fact that sine is periodic, hence it increases at one zero and decreases at the next, with the value of derivative $\pm 1$, while $(z-\pi k)$ always has derivative $1$ at $\pi k$. And so we obtain $$\frac{1}{\sin(z)}-\frac{1}{z}=\sum_{k=-\infty, k \ne 0}^{\infty} \frac{(-1)^{k}}{(1-\pi k z)}$$ $$\frac{1}{\sin(z)}=\sum_{k=-\infty}^{\infty} \frac{(-1)^{k}}{(1-\pi k z)}$$
Try to obtain the Mittag-Leffler decomposition for $\cot(z)$ as an exercise. Hint: it looks almost the same, and you can use the same system of contours.