TITLE: How do I find the series expansion of the meromorphic function $\frac{1}{e^z+1}$?
QUESTION [13 upvotes]: in a theoretical physics book, the author makes the following claim:
$$\frac{1}{e^z + 1} = \frac{1}{2} + \sum_{n=-\infty}^\infty \frac{1}{(2n+1) i\pi - z}$$
and justifies this as 

These series can be derived from a theorem which states that any meromorphic function may be expanded as a summation over its poles and residues at those poles

What's the name of that theorem? It's not really a Laurent series, since the Laurent series is for an expansion around one particular point only. I can see that the poles occur whenever $z = (2n+1)i\pi$ for $n \in \mathbb{N}$, but then where does that constant $1/2$ come from?
EDIT: Well, it appears that the general claim isn't valid, so now I'd be interested in a justification for the expansion in my particular example...

REPLY [4 votes]: The formula you quote can be easily obtained from the more general formula for the pole (or Mittag-Leffler) expansion of a meromorphic function $f$:
$$ \tag 1 f(z) = f(0) + \sum_{n=1}^\infty \alpha_n \left( \frac{1}{z-z_n} + \frac{1}{z_n} \right), $$
where

$f$ is a meromorphic function with simple poles at $z_n, \, n \in \mathbb{N}$,
holomorphic at $z=0$,
$\alpha_n$ is the residue of $f$ at the pole $z_n$,
$f$ is uniformly bounded on every circle $\mathcal C_n$ with radius $R_n$ containing all poles $z_k$ with $k \le n$, i.e.
$$ \tag 2\exists A > 0 | \,\, \forall (n \in \mathbb{N}, z \in \mathcal C_n), \,\, |f(z)| \le A  $$

Application of (1) to the problem at hand
Defining
$$f(z) = \frac{1}{e^z+1},$$
we see that $f$ has simple poles at $z_n = (2n+1)i \pi$ for $n \in \mathbb{Z}$, with residue $\alpha_n = -1$ for each $n \in \mathbb{N}$.
Direct application of (1), noting that $f(0)=1/2$, gives
$$ \frac{1}{e^z+1} = \frac{1}{2} - \sum_{n \in \mathbb{Z}} \left[ \frac{1}{z- i (2n+1) \pi} + \frac{1}{i(2n+1)\pi} \right].$$
The expansion given in the OP follows from the observation that
$$ \sum_{n \in \mathbb{Z}} \frac{1}{(2n+1)\pi} = 0. $$
Proof of (1)
Suppose without loss of generality that the poles $z_n$ are ordered monotonically as in
$$ 0 < |z_1| \le |z_2| \le \dots, $$
and fix some $z_0 \in \mathbb{C}$ such that $f$ is holomorphic at $z_0$.
Consider the integral
$$ \tag 3 I_n(z_0) \equiv \oint_{\mathcal C_n} \frac{dz}{2\pi i} \frac{f(z)}{z-z_0}, $$
which applying Residue's theorem becomes
$$ \tag 4 I_n(z_0) = f(z_0) + \sum_{k=1}^n \frac{\alpha_k}{z_k-z_0}. $$
In particular for $z_0=0$ we have
$$ \tag 5 I_n(0)
\equiv \oint_{\mathcal C_n} \frac{dz}{2\pi i} \frac{f(z)}{z}
= f(0) + \sum_{k=1}^n \frac{\alpha_k}{z_k}.$$
Taking the difference between (3) and (5) (in the integral form) we see that:
$$ \tag 6 |I_n(z_0) - I_n(0) | \le \oint_{\mathcal C_n} \frac{|dz|}{2\pi} \left| \frac{z_0 f(z)}{(z-z_0) z} \right| \le \frac{Az_0}{R_n - |z_0|} \to 0, \text{when } n \to \infty.$$
However, we also know that
$$ \tag 7  I_n(z_0)-I_n(0) = f(z_0) - f(0) - \sum_{k=1}^n \left( \frac{1}{z_0-z_k} + \frac{1}{z_k} \right), $$
which taking the limit $n\to \infty$ gives the wanted result (1).