On a parametrization of an infinite series to recover the GARCH process Time series analysis By James D. Hamilton (a great book) proceeds in this way to introduce the GARCH process:
First it recalls that the equation that described an ARCH(m) process was the following:
$$u_t = \sqrt{h_t} v_t$$
where $v_t$ is iid with zero mean and unit variance and where $h_t$ evolves according to
$$h_t = \xi + \alpha_1u_{t-1}^2+ \alpha_2u_{t-2}^2 + \dots+ \alpha_mu_{t-m}^2$$
Hamilton then invites us to imagine a more general process for wich the conditional variance depends on an infinite number of lags:
$$ h_t = \xi + \pi(L)u_t^2  $$
where $\pi(L) = \sum_{j = 1}^{\infty} \pi_j L^j$.
Hamilton then says that a natural idea is to parametrize $\pi(L)$ as the ratio of two finite polynomials as so
$$\pi(L) =\frac{ \alpha_1L^1+ \alpha_2L^2 + \dots + \alpha_mL^m}{1- \delta_1L^1-\delta_2L^2 - \dots -\delta_rL^r}$$
So why is this so natural?
 A: As this is, I argue, not a question about GARCH per se, I have added an ARMA tag.
Infinite order polynomials arise as soon as lag polynomials are inverted. 
Let me try to explain this via the more basic case of an ARMA process:
$$
Y_t=\phi_1Y_{t-1}+\phi_2Y_{t-2}+\ldots+\phi_pY_{t-p}+\epsilon_t+\;\theta_1\epsilon_{t-1}+\theta_2\epsilon_{t-2}+\ldots+\theta_q\epsilon_{t-q}
$$
In lag operator notation, this can be written more compactly as
$$
\phi(L)Y_t=\theta(L)\epsilon_t
$$
where
$$
\phi(L)=1-\phi_1L-\phi_2L^2-\ldots-\phi_pL^p
$$
and
$$
\theta(L)=1+\theta_1L+\theta_2L^2+\ldots+\theta_qL^q
$$
If the roots of $\phi(z)=0$ lie outside the unit circle (and if $\phi(z)$ and $\theta(z)$ have no common roots), we can invert the AR polynomial to find
$$
Y_t=\phi(L)^{-1}\theta(L)\epsilon_t=:\psi(L)\epsilon_t
$$
The $\psi_i$ can be found comparing coefficients: From $\phi(L)Y_t=\theta(L)\epsilon_t$ and $Y_t=\psi(L)\epsilon_{t}$ we have $$\phi(L)Y_t=\phi(L)\psi(L)\epsilon_t=\theta(L)\epsilon_t$$
and hence $\phi(L)\psi(L)=\theta(L)$.
Example: ARMA(1,1) 
We have
$$
(1-\phi L)(\psi_0+\psi_1L+\psi_2L^2+\psi_3L^3+\ldots)=1+\theta L
$$
and hence, by matching powers for a given power of $L$ on both sides of the equality,
\begin{align*}
\psi_0&=1\\
\psi_1-\phi\psi_0&=\theta\quad\Rightarrow\quad\psi_1=\phi+\theta\\
\psi_2-\phi\psi_1&=0\quad\Rightarrow\quad\psi_2=\phi(\phi+\theta)
\end{align*}
and in general, $\psi_j=\phi^{j-1}(\phi+\theta)$ for $j\geqslant 1$.
