Finding series representation of $\frac{1}{P(D)}$ through ordinary division

Question

I am studying ODEs from ordinary differential equations by Tenenbaum and Pollard. The book in its fifth chapter explains inverse operators for finding the particular solution of a constant coefficient linear ODE. It states that the particular solution $y_p = \frac{1}{P(D)}Q(x)$. It further goes on to prove this statement by using the following example:
$$ (D - a)y = bx^k $$ Which is equal to $$ y_p = \frac{1}{-a(1 - \frac{D}{a})}bx^{k} $$
It is further stated that $ \frac{1}{1 - \frac{D}{a}} = \sum\limits_{n=0}^{k}(\frac{D}{a})^n$ and states that this result was obtained by ordinary division. This is the part that confuses me. I do understand how this result could've been obtained by the usual Taylor series method but I do not understand how to obtain it from ordinary division (which I understand is polynomial long division) as the denominator is of a higher degree than the numerator. The book further goes on to state a similar result for a $n^{th}$ order linear constant coefficient ODE, which again according to the book was obtained through ordinary division. I would like to know how to use ordinary division to obtain a series representation for $\frac{1}{P(D)}$. Thank you.

Answer 1

The more useful perspectives, to my mind, are to note that this is the series expansion of $1/(1-x)$ at $x=0$ or to verify the expansion by applying $1-\frac Da$ to it and noting that the sum telescopes, the highest term vanishes and the lowest term is $1$.

However, if you really want to do this by "ordinary division" (I'm not sure how "ordinary" division can be when it involves differential operators), I'd do it like this (writing $x$ for $\frac Da$ to avoid unnecessary notational clutter):

What mutliple of $1-x$ do we need to get $1$? We need $1$, so the first term is $1$: $1/(1-x)=1+\ldots$. Now we have $1-x$, but we don't want the $x$, so what multiple of $1-x$ do we need to get rid of it? That would be $x$, so the next term is $x$: $1/(1-x)=1+x+\ldots$. Now we have $1-x^2$ but we don't want the $x^2$, so what multiple of $1-x$ do we need to get rid of it? That would be $x^2$, and so on...