Infinite series from Euler

Question

I am reading Euler's Introduction to analysis of the infinite" and is stumped at this section:

How does he derive the part that I circled in red?

I thought:

$\dfrac{a+bz+cz^2+...}{z(1-\alpha\cdot z-\beta\cdot z^2...)}=A+Bz+Cz^2+Dz^3+Ez^4...$

is gonna be:

$a+bz+cz^2+...=z(1-\alpha\cdot z-\beta\cdot z^2...)(A+Bz+Cz^2+Dz^3+Ez^4...)$

$=(z-\alpha\cdot z^2-\beta\cdot z^3...)(A+Bz+Cz^2+Dz^3+Ez^4)=(Az-\alpha\cdot Az^2-\beta\cdot Az^3...)+(Bz^2-\alpha\cdot B\cdot z^3-\beta\cdot B\cdot z^4...)$

So how does he have a fraction there, I don't understand.

Could you help me, please?

Answer 1

What you thought is impossible: the fraction has $0$ as a pole, and your r.h.s. has none. What our friend Leonard does is expanding first $\;\dfrac{a+b+cz^2+...}{1-\alpha\cdot z-\beta\cdot z^2 -\dotsb}$ via the standard formula for the geometric series, then multiplies the result by $\dfrac 1z$ (or $\dfrac1{z^m}$).

This way, he obtains an expansion of the function as what is known a Laurent series, since the function (and the series) has poles.

Answer 2

You are happy with geometrically expanding ? \begin{eqnarray*} \frac{1}{1- \alpha z - \beta z^2 -\cdots} &=& 1+(\alpha z + \beta z^2 -\cdots)+( \alpha z + \beta z^2 +\cdots)^2+ \cdots \\ &=& 1+pz+qz^2+ \cdots. \end{eqnarray*} Now let multiply in the numerator \begin{eqnarray*} \frac{a+bz+cz^2+\cdots }{1- \alpha z - \beta z^2 -\cdots} &=& (a+bz+cz^2+\cdots)( 1+pz+qz^2+ \cdots) \\ &=& A+Bz+Cz^2 + \cdots. \\ \end{eqnarray*} Now if we divide by $z$, we get \begin{eqnarray*} \frac{a+bz+cz^2+\cdots }{z(1- \alpha z - \beta z^2 -\cdots)} &=&&=& \frac{A}{z}+B+Cz + \cdots. \\ \end{eqnarray*} As required ?