= Solution
On any <compact set> avoiding the integers, write
$$
\frac1{z-n}+\frac1n=\frac{z}{n(z-n)}.
$$
For all sufficiently large $|n|$, uniformly on that <compact set> this is $O(|n|^{-2})$. The summands in the higher series are $O(|n|^{-k})$ for $k\ge2$. The <Weierstrass M-test> therefore gives <locally uniform convergence> away from the integer <poles> and permits termwise differentiation. Near an integer $m$, isolate its singular summand; all the others converge uniformly on a small <neighborhood>. Thus \b[the first function has a simple <pole> of <residue> one at each integer, and the $k$th function has a <pole> of order $k$ there with principal part $(z-m)^{-k}$]. There are no other singularities, so all are <meromorphic functions>.
Pairing positive and negative indices is legitimate because the regularized first series converges absolutely. It becomes
$$
\varepsilon_1(z)=\frac1z+\sum_{n=1}^{\infty}\frac{2z}{z^2-n^2}.
$$
Here is a direct derivation of its <cotangent partial-fraction expansion>. Fix $z\notin\mathbb Z$ and integrate $F(\zeta)=\pi\cot(\pi\zeta)/(\zeta^2-z^2)$ around the square with real and imaginary coordinates $\pm R$, $R=N+1/2$, taking $N$ large enough to enclose $\pm z$. On the vertical sides $|\cot\pi\zeta|\le1$, and on the horizontal sides it is at most $\coth(\pi R)$. Thus the <cotangent> is uniformly bounded and $|\zeta^2-z^2|\ge R^2-|z|^2$, so the contour integral is $O(R^{-1})\to0$.
The <residues> at integers are $1/(n^2-z^2)$; the two <residues> at $z$ and $-z$ sum to $\pi\cot(\pi z)/z$. The <residue theorem> and <absolute convergence> of the integer <residue> sum consequently give
$$
0=\frac{\pi\cot\pi z}{z}+\sum_{n\in\mathbb Z}\frac1{n^2-z^2},
\qquad\varepsilon_1(z)=\pi\cot\pi z.
$$
Using the exponential definitions of <sine> and <cosine> now yields, for $w=e^{2\pi iz}$,
$$
\boxed{\varphi_1(w)=\pi i\frac{w+1}{w-1},\qquad
\varepsilon_1(z)=\varphi_1(e^{2\pi iz}).}
$$
The same <locally uniform convergence> gives $\varepsilon_k'=-k\varepsilon_{k+1}$, including $k=1$. Hence if $\varepsilon_k(z)=\varphi_k(e^{2\pi iz})$, the <chain rule> gives
$$
\boxed{\varphi_{k+1}(w)=-\frac{2\pi i}{k}\,w\varphi_k'(w).}
$$
This is a <rational function> whenever $\varphi_k$ is, proving the requested descent for every $k$ by induction, with
$$
\boxed{\varphi_2(w)=-\frac{4\pi^2w}{(w-1)^2}.}
$$
In particular periodicity is a consequence of the explicit descent, rather than an assumption about the summation order.
Finally, for $|z|<1$ expand each paired summand as a <geometric series>:
$$
\frac{2z}{z^2-n^2}=-2\sum_{j=0}^{\infty}\frac{z^{2j+1}}{n^{2j+2}}.
$$
On $|z|\le r<1$ the double series converges absolutely, since its absolute sum is at most $2r\sum_{n\ge1}n^{-2}/(1-r^2)$. Interchanging the sums and using the <Riemann zeta function> gives the <Laurent series>
$$
\boxed{\varepsilon_1(z)=\frac1z-2\sum_{j=0}^{\infty}\zeta(2j+2)z^{2j+1},\qquad0<|z|<1.}
$$
Thus the $z^{-1}$ coefficient is one; the positive odd coefficients are $-2\zeta(2j+2)$, and all other coefficients vanish.
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