Conditions for the summation theorem for functions with infinitely many poles

Question

It is widely know the summation theorem via residues:

Suppose $f(z)$ analytic on $\mathbb{C}$ except for a finite number of poles. And let $f(z)$ be such that along the path $C_{N}$, the square with vertices at $\left(N+\frac{1}{2}\right)(-1+i), \left(N+\frac{1}{2}\right)(-1-i), \left(N+\frac{1}{2}\right)(1-i), \left(N+\frac{1}{2}\right)(1+i)$.

$$|f(z)|\leq\frac{M}{|z|^k}$$

where $k>1$ and $M$ are constants independent of $N$ we have

$$\sum_{n=-\infty}^{\infty} f(n) = -\sum \left\{\textrm{ residues of } \pi\cot(\pi z)f(z) \textrm{ at the poles of } f(z) \right\}$$

We can extend this theorem for functions with infinitely many poles:

Consider the path $C_{N}$. Let $f(z)$ with infinitely many poles with the conditions:

$$|f(z) |\leq \frac{M}{|z|^{k}} \tag{*}$$ along the path $C_{N}$ where $k>1$ and $M$ are constants independent of $N$ and

$$\lim_{N \to \infty} \sum_{a\in A_{N}} \operatorname{Res}\left(\pi\cot(\pi z)f(z),a\right)<\infty$$

where

$$A_{N} = \left\{ \textrm{poles of } f(z) \textrm{ inside } C_{N} \right\}$$

then:

$$\sum_{n=-\infty}^{\infty} f(n) = \lim_{N \to \infty} \sum_{a\in A_{N}} \operatorname{Res}\left(\pi\cot(\pi z)f(z),a\right)$$

the set of $C_{N}$ is called "Big squares".

1. Can we weaken the condition $(*)$? For example, $f(z)$ to be $\mathcal{O}(z^{-2})$ at infinity?

$$|f(z)| \leq \frac{M}{|z|^{2}} \quad z \to \infty$$ or there are even weaker conditions?

2. What is a good reference for the summation theorem for functions with infinitely many poles?

Here is my try for question 1:

Given that $f(z)$ is $\mathcal{O}(z^{-2})$ :

$\exists \delta,M>0$ so that whenever $|z|>\delta$

$$|f(z)|\leq\frac{M}{|z|^2}$$

If we choose our first square $C_{N}$ for $N$ sufficiently large such that $|z|>\delta$ then

$$\oint_{C_{N}} \pi \cot(\pi z)f(z) dz = \sum_{n=-N}^{N} f(n) + \sum_{a\in A_{N}} \operatorname{Res}\left(\pi\cot(\pi z)f(z),a\right) $$

We just have to prove that

$$\oint_{C_{N}} \pi \cot(\pi z)f(z) dz \to 0 \quad N \to \infty$$

But it can be proven that $$|\cot(\pi z)|$$ is bounded in $C_{N}$:

$$|\cot(\pi z)|\leq A$$

Then

$$\left| \oint_{C_{N}} \pi \cot(\pi z) f(z) dz\right| \leq \frac{\pi A M}{N^2}(8N+4)$$

since the lenght of the path $C_{N}$ is $8N+4$.

Hence, the integral vanishes as $N \to \infty$.

Answer 1

EDIT: I reorganized my answer.

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The issue is that a meromorphic function with infinitely many poles can't be $O(z^{-k}), \ k >1, $ as $|z| \to \infty$ since if would require the poles of the function to all be located inside a circle of finite radius centered at the origin.

The poles would invariabily accumulate near some point, contradicting the fact that $f(z)$ only has isolated singularities.

It would be better to write $f(z)$ as the product $f(z) = g(z)h(z)$, where $g(z)$ is a meromoprhic function with infintiely many poles, and $h(z)$ is $O(z^{-k}), \ k>1$, as $|z| \to \infty$.

You would then need to argue that $g(z)$ is uniformly bounded on the contour like $\cot(\pi z)$.

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If you stay away from the negative real axis, then the polygamma function $\psi^{(2)}(z)$ is a meromoprhic function with infinitely many poles that is $O(z^{-2})$ as $|z| \to \infty$.

However, you would need to examine how $\psi^{2}(z)$ behaves on the part of the contour that passes halfway between adjacent poles on the negative real axis as $N$ increases.

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And when using the kernel $\pi \csc(\pi z)$ to evaluate an alternating series, you only need $h(z)$ to be $O(z^{-k}), \ k>0,$ as $|z| \to \infty$.

Basically it has to do with the fact that the magnitude of $\csc(\pi z)$ decays exponentially to zero as $\Im(z) \to \pm \infty$. See the answers to this question.

And depending on how $g(z)$ behaves as $\Re(z) \to \pm \infty$, $h(z) =1$ might suffice.