I have two questions about my two different attempts at the following exercise.
Question 1: Does the first method = attempt 1 not work because we can concretely determine what each term of the Laurent series looks like? For example, we have infinitely many constant terms when we try to multiply our the factors.
Question 2: Is it because $f(z)$ is centered around $0$ that we are able to use the normal Taylor expansion of $\frac{1}{1-z}=\sum z^n$? If we kept the annulus as $1<z<2$ still but changed the center of $f(z)$, can we still use the Taylor expansion of $\frac{1}{1-z}$ in Attempt 2?
Exercise: Write the three terms of the Laurent expansion of $f(z)=\frac{1}{z(z-1)(z-2)}$ centered at $0$ and convergent in $1<|z|<2$.
Attempt 1 (naive method): $\frac{1}{z-1}=\frac{1}{z}\sum_{n=1}^{\infty}\left(\frac{1}{z}\right)^n$ and $\frac{1}{z-2}=\frac{1}{2(\frac{z}{2}-1)}=\frac{-1/2}{1-\frac{z}{2}}=-\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{z}{2}\right)^n$. I divided 1 by $z$ and $z$ by 2 so that $|1/z|<1$ and $|z/2|<1$ for the respective Taylor series to converge.
Then multiplying $f(z)$ out we have $$f(z)=-\frac{1}{2}\frac{1}{z}\left(\frac{1}{z}\sum_{n=1}^{\infty}\left(\frac{1}{z}\right)^n\right)\left(\sum_{k=1}^{\infty}\left(\frac{z}{2}\right)^k\right)=-\frac{1}{2}\frac{1}{z}\left(\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+\cdots\right)\left(\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+\cdots\right).$$
Attempt 2 (correct method): I start by doing partial decomposition.$$\frac{1}{z(z-1)(z-2)}=\frac{A}{z}+\frac{B}{z-1}+\frac{C}{z-2}$$ yields $A=1/2, B=-1, C=1/2$.
Then applying the Taylor series from above, we have $$f(z)=\frac{1}{2z}-\frac{1}{z}\sum_{n=1}^{\infty}\left(\frac{1}{z}\right)^n-\frac{1}{4}\sum_{k=1}^{\infty}\left(\frac{z}{2}\right)^k= -\frac{1}{2}z^{-1}-\frac{1}{4}-\left(\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}+\cdots\right) -\left(\frac{z}{8}+\frac{z}{16}+\frac{z^3}{32}+\cdots\right).$$
So the first three nonnegative terms of the Laurent series are $-1/4, -z/8, -z^2/16$.