Laurent Series of a Rational Polynomial

I am trying to compute the Laurent expansion of $$\frac{x}{x-1}$$ around $$\alpha=2$$. I wrote it out as a partial fraction expansion so we have: $$\frac{x}{x-1} = 1 + \frac{1}{x-1} = 1-\frac{1}{1-x}$$ And so we should have the Laurent expansion of this by the geometric series obtaining: $$1-\sum_{n=0}^\infty (x-2)^n = -(x-2)-(x-2)^2-\cdots$$ But this is wrong, according to the textbook I am working with the correct answer should be: $$2-(x-2)+(x-2)^2-(x-2)^3+\cdots$$ I don't know where exactly I've gone wrong either, I suspect I've made a mistake in computing the expansion at $$\alpha=2$$ because I know that when $$\alpha=0$$ this would have given the correct result. I'd just like to understand where I went wrong, I think I am misunderstanding something critical.

• Do you expect us to guess where you made a mistake ? – Yves Daoust May 13 at 13:14
• @YvesDaoust Sorry if the question wasn't formulated better. I thought it was clear that I had a misunderstanding on how exactly we move from computing a Laurent series expansion at $\alpha=0$ to an arbitrary point. I believe that was where my mistake was, but I didn't understand where the method I used didn't work... seemed like a reasonable enough question to me. – Logan Toll May 13 at 13:17

$$1+\dfrac1{x-1} =1+\dfrac1{1+(x-2)}=1+1-(x-2)+(x-2)^2-(x-2)^3+...$$
You went wrong when you wrote that$$\frac1{1-x}=\sum_{n=0}^\infty(x-2)^n.$$The sum of this series is actually$$\frac1{1-(x-2)}=\frac1{3-x}.$$