Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$

Question

I am trying to find the Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$ when $0<|z-1|<1$.

I thought that if \begin{align} f(z)&=\frac{1}{z-1}-\frac{1}{z-2}+\frac{1}{(z-2)^2} \\ &=\frac{1}{z-1}+\frac{1}{2-z}+\frac{d}{dz}\left(\frac{1}{2-z}\right) \\ &=\frac{1}{z-1}+\frac{1}{1-(z-1)}+\frac{d}{dz}\left(\frac{1}{1-(z-1)}\right) \\ &=\frac{1}{z-1}+\sum_{n=0}^{\infty}(z-1)^n+\frac{d}{dz}\left(\sum_{n=0}^{\infty}(z-1)^n\right) \\ &=\frac{1}{z-1}+\sum_{n=0}^{\infty}\left((z-1)^n+n(z-1)^{n-1}\right) \\ &=\frac{1}{z-1}+\sum_{n=0}^{\infty}(z-1)^n\left(1+n(z-1)^{-1}\right) \\ \end{align} But this does not agree with the series generated by wolfram. What about this method is incorrect?

alternative approach suggested by David

\begin{align} f(z)&=\frac{1}{z-1}\left(\frac{1}{(z-2)^2}\right) \\ &=\frac{1}{z-1}\frac{d}{dz}\left(-\frac{1}{z-2}\right) \\ &=\frac{1}{z-1}\frac{d}{dz}\left(\frac{1}{1-(z-1)}\right) \\ &=\frac{1}{z-1}\frac{d}{dz}\left(\sum_{n=1}^{\infty}(z-1)^n\right) \\ &=\frac{1}{z-1}\left(\sum_{n=1}^{\infty}n(z-1)^{n-1}\right) \\ &=\sum_{n=1}^{\infty}n(z-1)^{n-2} \end{align}

Answer 1

1. In line 2, $-\frac{d}{dz}$ should be $+\frac{d}{dz}$.
2. Your final sum has many powers duplicated and these could be collected into single terms. For example when $n=5$ the first term is $(z-1)^5$ and when $n=6$ the second term is $6(z-1)^5$.
3. Easier method: take $\frac1{z-1}$ as a multiplicative factor at the beginning.

Answer 2

Use this one: let $z-1=w$ to make it easier then with $0<|w|<1$ $$\dfrac{1}{1-w}=\sum_{n\geq0}^\infty w^n$$ differentiating shows $$\dfrac{1}{(1-w)^2}=\sum_{n\geq1}^\infty nw^{n-1}$$ therefore $$\dfrac{1}{w(w-1)^2}=\sum_{n\geq1}^\infty nw^{n-2}$$