I've a doubt about this exercise
Find the Laurent series of the function $$f(z) = \frac{1}{z(z-2)^2}$$ addend $z=0$ in the annular regions of interest
The problem it's not totally about the how to evaluate it, but instead about the fact that using two different methods I get two different Laurent series. Let me explain my thought.
First my question is only about the "first" region $|z|<2$.
Fist method
With some algebraic manipulations on the series I get that $$f(z) = \frac{1}{z}\frac{1}{(z-2)^2} = \frac{1}{z}\frac{d}{dz}\left(\frac{-1}{z-2}\right) = \frac{1}{z}\frac{d}{dz}\left(\frac{1}{2-z}\right)\\ = \frac{1}{z}\frac{d}{dz}\left(\frac{1}{2(1-\frac{z}{2})}\right) = \frac{1}{2z}\frac{d}{dz}\color{red}{\left(\frac{1}{1-\frac{z}{2}}\right)}$$ The red factor it's just a geometric series because $$|z|\lt 2 \implies \left|\frac{z}{2}\right|\lt 1$$ Using this fact what I get is $$f(z) = \frac{1}{2z}\frac{d}{dz}\sum_{n=0}^\infty\left({z\over 2}\right)^n$$ bringing the derivative into the sum I finally get the result $$f(z) = \frac{1}{2z}\sum_{n=0}^\infty\frac{d}{dz}\left({z\over 2}\right)^n = \frac{1}{4z}\sum_{n=1}^\infty n\left({z\over 2}\right)^{n-1} = \sum_{n=1}^\infty \frac{nz^{n-1}}{2^{n+1}}$$ and this, according to Wolfram, is the right result.
Second method
Taking the derivation from the third step of the method before I've done different algebraic manipulations, mainly $$f(z) = \frac{1}{z}\frac{d}{dz}\left(\frac{1}{2-z}\right) = \frac{1}{z}\frac{d}{dz}\left(\frac{1}{2-z\color{blue}{+1-1}}\right) = \frac{1}{z}\frac{d}{dz}\color{red}{\left(\frac{1}{1-(z-1)}\right)}$$ again the red factor should, I explain in a moment, be a geometric series because $$|z|\lt 2\implies z-1\lt 1\tag 1$$ taking this for good I continue with the calculations $$f(z) = \frac{1}{z}\frac{d}{dz}\sum_{n=0}^\infty (z-1)^n = \frac{1}{z}\sum_{n=0}^\infty \frac{d}{dz}(z-1)^n = \frac{1}{z}\sum_{n=1}^\infty n(z-1)^{n-1}$$ I'm stopping here because this doesn't seem, to me, the same as before!
The questions then are:
Question 1: Why are they not the same? Where is the wrong assumption?
Question 2: Is it wrong to assume $(1)$?
Question 3: If there's nothing wrong, which I doubt it, how can I go from the second series to the first with some algebraic manipulations?