Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ Consider $$f(z)=\frac{1}{z(z-1)(z-2)}.$$
I want to determine the Laurent series in the point $z_0=0$ on $0<|z|<1$.
Partial decomposition yields:
$$f(z)=\frac{1}{z(z-1)(z-2)}=(1/2)\cdot (1/z) - (1/(z-1)) + (1/2)(1/(z-2)).$$
Is the general strategy now, to try to use the geometric series?
$(1/(z-1))=-(1/(1-z))=-\sum_{k=0}^\infty z^k$
$\displaystyle 1/(z-2)=-(1/2)\frac{1}{1-\frac{z}{2}}=-(1/2)\cdot\sum_{k=0}^\infty (z/2)^k$
So $f(z)=(1/2)\cdot (1/z)+(1/2)\cdot\sum_{k=0}^\infty z^k-(1/4)\cdot\sum_{k=0}^\infty (z/2)^k$ (*)
Some questions:
1) What is the difference between a Laurent and a Taylor series? I don't get it. It seems you calculate them the same.
2) Why didn't we write (1/z) as a series expansion too?
3) What makes the end result (*) to be a Laurent series?
 A: Partial fractions and geometric series give
$$
\begin{align}
\frac1{(1-x)(2-x)}
&=\frac1{1-x}-\frac1{2-x}\\
&=\frac1{1-x}-\frac12\frac1{1-x/2}\\
&=(1+x+x^2+x^3+x^4+\dots)\\
&-\left(\frac12+\frac14x+\frac18x^2+\frac1{16}x^3+\frac1{32x^4}+\dots\right)\\
&=\frac12+\frac34x+\frac78x^2+\frac{15}{16}x^3+\frac{31}{32}x^4+\dots
\end{align}
$$
Thus, the Laurent series for $\frac1{x(x-1)(x-2)}$ at $x=0$ is
$$
\frac1{x(x-1)(x-2)}=\frac1{2x}+\frac34+\frac78x+\frac{15}{16}x^2+\frac{31}{32}x^3+\dots
$$
We could also expand the series at $x=1$. Let $y=x-1$ and then
$$
\begin{align}
\frac1{x(x-1)(x-2)}
&=\frac1{(y+1)y(y-1)}\\
&=-\frac1y\frac1{1-y^2}\\
&=-\frac1y-y-y^3-y^5-y^7-\dots\\
&=-\frac1{x-1}-(x-1)-(x-1)^3-(x-1)^5-(x-1)^7-\dots
\end{align}
$$


*

*The Laurent series is much like the Taylor series except terms of negative degree are allowed.

*We don't expand $\frac1x$ since there is no power series for $\frac1x$ at $0$ other than $\frac1x$.

*Definition. It is a power series at a point, $x_0$ which can have both positive and negative powers of $x-x_0$.
