ARCH(∞) = GARCH(p,q) proof I am aware that the similiar question was asked already here. However, I read Bollerslev (1986) and I struggle hard with the rearrangements and substitutions he makes. Hence, there are subquestions to the one in the title that are directly linked to the steps of the proof I will outline below. 
First, the initial equation is 
$$ h_t = \alpha_0 + A(L)\epsilon^2_t + B(L)h_t \qquad (1)$$ 
where $A(L)$ and $B(L)$ denote some kind of backwards operator. 
Then, he rearranges the equation under the condition that all roots of 1 - B(z) = 0 lie outside the unit circle (Why?) to 
$$ h_t = \frac{a_0}{1 - B(1)} + \frac{A(L)}{1 - B(L))}\epsilon^2_t $$ 
Obviously the right-hand term in (1) is now dropped. But why? I really don't follow. But it gets even more inapprehensible for me when next 
$$ h_t = \frac{a_0}{1-\sum_{i=1}^{p} \beta_i} + \sum_{i=1}^{∞} d_i\epsilon^2_{t-i} \qquad (2) $$
My question about (2) is: why is $1 - B(1) = 1-\sum_{i=1}^{p} \beta_i$ instead of $ 1 - \beta_1$? 
The last subquestion concerns the power series expansion of $D(L) = \frac{A(L)}{1 - B(L)}$ to 
$$ d_i = \alpha_i + \sum \limits_{j=1}^{n} \beta_j d_{i-j},   \qquad i = 1, ..., q  \qquad (3)$$ which is equivalent to 
$$ d_i = \sum \limits_{j=1}^{n} \beta_j d_{i-j},  \qquad   i = q+1, ...  \qquad (4).$$
How do equations (3) and (4) come into existence? I am wildly confused. Maybe somebody possesses clairvoyance and can shed some light on this? 
 A: $h_t = \alpha_0 + A(L)\epsilon^2_t + B(L)h_t$
$h_t -  B(L)h_t = \alpha_0 + A(L)\epsilon^2_t $
$[1 -  B(L)]h_t = \alpha_0 + A(L)\epsilon^2_t $
$h_t = [1 -  B(L)]^{-1}(\alpha_0 + A(L)\epsilon^2_t) $
(as long as you can take that inverse... and this is where the roots outside unit circle condition you ask about comes in handy)
$\quad = [1 -  B(L)]^{-1}\alpha_0 + [1 -  B(L)]^{-1}A(L)\epsilon^2_t) $
Now this can be written in shorthand as your second equation apart from the typo in your first term.
Note that A(L) is of finite length, but if you expand $[1-B(L)]^{-1}$ as a power series in $L$, it will be an infinite series. You want that series to converge, whence the restriction on the roots.
So now taking $D(L)=[1-B(L)]^{-1}A(L)$ note that you have 
$[1-B(L)]D(L)=A(L)$
So $D(L)  = A(L)+ B(L)D(L)$
Now expand out and equate term by term (i.e. the coefficient of $L$ on the LHS must be the coefficient of $L$ on the RHS, then the coefficient of $L^2$ and so on)
Notice that up to the term in $L^q$ there's an $\alpha$ in there but after that it drops out, which is why after that point you look at equation 4 instead of equation 3.
A: whuber told me that it is okay to answer my own questions. I am not sure if it is still okay for subquestions. Nevertheless, I found out that (1) is transformed into 
\begin{align} 
\frac{h_t}{1 - B(L)} &= \frac{a_0}{1 - B(1)} + \frac{A(L)}{1 - B(L)}\epsilon^2_t + \frac{B(L)h_t}{1 - B(L)} \\ 
\\
\frac{h_t}{1 - B(L)} - \frac{B(L)h_t}{1 - B(L)} &= \frac{a_0}{1 - B(1)} + \frac{A(L)}{1 - B(L)}\epsilon^2_t \\
\\
h_t &= \frac{a_0}{1 - B(1)} + \frac{A(L)}{1 - B(L)}\epsilon^2_t
\end{align}
this way. Also, I am now pretty confident that $\frac{a_0}{1 - B(1)}$ has to be a typo with the right one being $\frac{a_0}{1 - B(L)}$. What I am still trying to figure out is the last part, that is this mysterious power series expansion. 
(If this answer to the subquestion is inappropriate I will put it in an edit.)
