<s>[INST] The Residual of $\frac{1}{(z-1)(z+2)z}$ I am calculating the residual of  $f(z)=\frac{1}{(z-1)(z+2)z}$ on $z=1$
Since it is a first order pole, It is easy to know that $Res(z=1)=\lim_{z\to1}(z-1)f(z)=\frac{1}{3}$
But when I calculate the Laurent Series:
$f(z)=\frac{1}{(z-1)(z+2)z}=\frac{1}{z-1}(\frac{1}{z}-\frac{1}{z+2})\frac{1}{2}$
and $\frac{1}{z}=\frac{1}{(z-1)+1}=\frac{1}{z-1}\frac{1}{1+\frac{1}{z-1}}=\frac{1}{z-1}\{1-\frac{1}{z-1}+\frac{1}{(z-1)^2}+\dots\}$
$\frac{1}{z+2}=\frac{1}{(z-1)+3}=\frac{1}{3}\frac{1}{1+\frac{z-1}{3}}=\frac{1}{3}\{1-\frac{z-1}{3}+\frac{(z-1)^2}{3^2}-\dots\}$
Therefore:$f(z)=\frac{1}{2}\{\frac{1}{(z-1)^2}-\frac{1}{(z-1)^3}+\dots\}-\frac{1}{6}\{\frac{1}{z-1}-\frac{1}{3}+\frac{z-1}{3^2}-\dots\}$
Then the coefficient of $\frac{1}{z-1}$ is $-\frac{1}{6}$. I remember that the residual of $z=z_0$ is actually the the coefficient $a_{-1}$ in the Laurent Series $f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$. What mistakes did I make?
 [/INST] The series you wrote for $\frac 1 z$ is valid only when $|\frac  1 {z-1}| <1$ or $|z-1| >1$. The correct expansion of $\frac  1z$ is $1-(z-1)+(z-1)^{2}+\cdots$ valid for $|z-1|<1$.
</s>