<s>[INST] Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$ 
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}
 [/INST] *

*In line 2, $-\frac{d}{dz}$ should be $+\frac{d}{dz}$.

*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 $6(z-1)^5$.

*Easier method: take $\frac1{z-1}$ as a multiplicative factor at the beginning.

</s>