Find a power series by comparing it to a geometric series?

Question

Find the power series for the following function: $f(x)=\frac{1-x}{x-3}$ centered at x=1.

This is what I've done: $$\frac{1-x}{x-3}=(1-x)\sum^{\infty}_{n=0}{(2-x-1)^n}=(1-x)\sum^{\infty}_{n=0}{(1-x)^n}=(1-x)\sum^{\infty}_{n=0}{(-1)^n(x-1)^n}$$ which converges since $$\lim_{n\to\infty}{|\frac{(-1)^n}{(-1)^{n+1}}|} = 1$$ However, when I compute the coefficients by hand using $a_k=\frac{f^{k}(x_o)}{k!}$, where the series is centered at 1, I get different values for a_k making me think the series is wrong, but I'm not sure where my error is.

Answer 1

An idea:

$$\frac{1-x}{x-3}=(x-1)\frac12\frac1{1-\frac{x-1}2}=\frac12(x-1)\sum_{n=0}^\infty\left(\frac{x-1}2\right)^n=\sum_{n=1}^\infty\frac{(x-1)^n}{2^n}$$

Answer 2

I think you got a sign wrong. You have got terms in $(2-x-1)^n=(1-x)^n$ but the denominator is $-2+(x-1)=-2-(1-x)$.