How do I find the series expansion of the meromorphic function $\frac{1}{e^z+1}$?

Solution 1:

Mittag-Leffler's theorem guarantees the existence of a meromorphic function $g(z)$ whose poles and principal parts are given by any values specified. Then, if $f(z)$ is a meromorphic function, then $f(z) - g(z)$ is holomorphic, and it remains to compute this difference. In practice this is probably nontrivial, because $g(z)$ is not uniquely determined, but for functions with nice poles and principal parts, this is possible.

Such a possibility applies in your case with $f(z) = 1/(e^z + 1)$. We can justify the formula you gave in your question by using an approach based on a discussion between me and one of my friends, so I do not claim the credit for these ideas.

In order to properly handle the convergence of the infinite sum, we should first symmetrize the infinite sum you gave, so instead let $$ g(z) = \sum_{k > 0 \text{, odd}} \left( \frac{1}{k i \pi - z} - \frac{1}{k i \pi + z} \right) = -\sum_{k > 0 \text{, odd}} \frac{2z}{z^2 + k^2 \pi^2}$$

We can check that $g(z)$ is a meromorphic function whose poles and principal parts match that of $f(z) = 1/(e^z + 1)$, so it follows that $f(z)- g(z)$ is entire. It remains to compute this difference. First, notice that both $f(z)$ and $g(z)$ are $2 \pi i$ periodic. So to check the growth of $f(z), g(z)$, we need only check the behavior as $\mathrm{Re}(z) \rightarrow \pm \infty$. Notice that $f(z) \rightarrow 1,0$ as $\mathrm{Re}(z) \rightarrow -\infty, +\infty$, respectively. Thus it follows that $f(z)$ is in fact uniformly bounded away from its poles. To check $g(z)$, split the sum as $$ g(z) = \sum_{0<k<2|z|/\pi, \text{ odd}} + \sum_{k \ge 2|z|/\pi, \text{ odd}} \frac{-2z}{z^2 + k^2 \pi^2} = S_1(z) + S_2(z)$$ Now, notice that for $\mathrm{Re}(z)$ sufficiently large, \begin{align} |S_1(z)| & = \left|\frac{-2}{z} \sum_{0<k<2|z|/\pi, \text{ odd}} \frac{1}{1 + k^2 \pi^2/z^2} \right| \le \frac{2}{|z|} \frac{2|z|}{\pi} = 4,\\ |S_2(z)| & = \left| \sum_{k \ge 2|z|/\pi, \text{ odd}} \frac{-2z}{z^2 + k^2 \pi^2} \right| \\ & \le \frac{8}{3} |z| \sum_{k \ge 2|z|/\pi, \text{ odd}} \frac{1}{ \pi^2 k^2} \\ & \le \frac{8}{3} |z| \int_{-1 + 2|z|/\pi}^{\infty} \frac{1}{\pi^2 s^2} \, ds \\ & \le \frac{8}{3} \frac{|z|}{-1 + 2|z|/\pi} \le C \end{align} for $C > 0$ a constant. Thus $g(z)$ is also uniformly bounded away from its poles. Then, it follows that the difference $f(z) - g(z)$ is uniformly bounded, and being entire, then by Louiville's theorem it must be constant. Now, we compute that $$ f(0) - g(0) = \frac{1}{2} - 0 = \frac{1}{2} $$ and hence $f(z) - g(z) \equiv 1/2$.

Solution 2:

Taken literally, the claim is certainly not true. Meromorphic functions are not uniquely determined by their poles and corresponding residues. Take for example $$ \frac{1}{z} \qquad\text{and}\qquad \frac{e^z}{z}, $$ which have the same poles and same residues.

However, Mittag-Leffler's theorem as noted by Random Variable is related.

For your particular example, you can start from a well known series expansion of $\cot z$: $$\cot z = \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{z+n\pi} = \frac1z + \sum_{n=0}^\infty \frac{2z}{n^2\pi^2-z^2},$$ see for example this for a derivation.

Let's interpret your series as a symmetric sum and rewrite it by summing the terms pairwise: $$(\operatorname{term} 0 + \operatorname{term} -1) + (\operatorname{term} 1 + \operatorname{term} -2) + \cdots $$

This will give us $$ \sum_{n=-N-1}^N \frac{1}{(2n+1)i\pi - z} = -i\sum_{n=0}^N \frac{2z}{(2n+1)^2\pi^2 + z^2}, $$ which is very similar to the series for $\cot z$, but where we have thrown away half the terms. Inspired by this observation, let's compute $$ \cot z - \frac12\cot \frac z2 = \frac1z + \sum_{n=0}^\infty \frac{2z}{n^2\pi^2-z^2} - \frac1{z} - \sum_{n=0}^\infty \frac{2z}{4n^2\pi^2-z^2} = \sum_{n=0}^\infty \frac{2z}{(2n+1)^2\pi^2-z^2}. $$ Hence $$ \cot iz - \frac12\cot \frac {iz}{2} = \sum_{n=0}^\infty \frac{2z}{(2n+1)^2-(iz)^2} = \sum_{n=0}^\infty \frac{2z}{(2n+1)^2\pi^2+z^2}. $$ Consequently, your sum is \begin{align} \frac12 -i\sum_{n=0}^\infty \frac{2z}{(2n+1)^2\pi^2 + z^2} &= \frac12+\frac i2\cot \frac {iz}{2} - i\cot iz \\ &= \frac{1}{e^z+1}. \end{align}

Solution 3:

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

1. $f$ is a meromorphic function with simple poles at $z_n, \, n \in \mathbb{N}$, holomorphic at $z=0$,
2. $\alpha_n$ is the residue of $f$ at the pole $z_n$,
3. $f$ 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).