Pole expansion using Mittag-Leffler's theorem

Question

In Arfken's Mathematical Methods for Physicists, we have the following statement of a result due to Mittag-Leffler. We assume there are poles at $0<|a_1|<|a_2|<...$ with residues $b_n$.

"Let us consider a series of concentric circles $C_n$ about the origin so that $C_n$ includes $a_1,a_2,...a_n$ but no other poles, its radius $R_n \rightarrow \infty$ as $n\rightarrow\infty$. To guarantee convergence we assume that $|f(z)|<\epsilon R_n$ for any small positive constant $\epsilon$ and all z on $C_n$. Then the series $$f(z)=f(0)+\sum_{n=1}^{\infty} b_n(\left(z-a_n\right)^{-1}+a_n^{-1})$$ converges to $f(z)$."

I'm confused by the $\epsilon$. For which $n$ does the function satisfy this bound? If the function satisfies that bound for each $n$ and any $\epsilon$, then the requirement sounds much too specific for the theorem to be of any use - $f$ would need to be zero on each $C_n$ to satisfy that bound, and that's not realistic to expect for a reasonable function.

I'm sure there's a communication breakdown between Arfken and I. Can someone help resolve this?

Answer 1

If $f$ is meromorphic on $\Bbb{C}$ with only simple poles at $(a_k)\in \Bbb{C}^*$ of residues $(b_k)$ and there is a sequence $R_n\to \infty$ such that $$\lim_{n \to \infty} \frac{\sup_{|z|= R_n}|f(z)|}{R_n} = 0$$ then the residue theorem gives

$$0= \lim_{n \to \infty}\frac1{2i\pi}\int_{|s|=R_n} \frac{-zf(s)}{s(s-z)}ds=\lim_{n \to \infty}\frac1{2i\pi}\int_{|s|=R_n} f(s)\left(\frac1s-\frac1{s-z}\right)ds$$ $$= \lim_{n \to \infty}f(0)-f(z)+\sum_{|a_k|< R_n} b_k \left(\frac1{a_k}-\frac{1}{a_k-z}\right)$$

An example is $f(z) = \dfrac1{e^{2i\pi z}+1}, a_k = k+\dfrac12, b_k=\dfrac{1}{2i\pi},R_n=n$

To ensure the series converges in usual sense once the $a_k$ have been ordered by magnitude we need some additional conditions on the number of poles on each annulus $(R_n,R_{n+1})$ and on the magnitude of the residues.