<s>[INST] How to express $f(x)=\frac{1}{x-1} + \frac{1}{x-3}$ as a power series? I am working through my Calculus 2 course problems, and I have reached..

Express the function $f(x)=\frac{2x-4}{x^2-4x+3}$ as the sum of a power series by first using partial fractions.

I can do the partial fractions..
$$\frac{2x-4}{x^2-4x+3}=\frac{2(x-2)}{(x-1)(x-3)} =\frac{A}{x-1}+\frac{B}{x-3}$$
$$2(x-2)=A(x-3)+B(x-1)$$
$$\to -2=-2A \to1=A$$
$$\to2=2B\to 1 = B$$
$$\frac{1}{x-1} + \frac{1}{x-3}$$
I am not sure how to approach creating a power-series with two fractions like this. 
 [/INST] Separately,
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 +\> ...$$
$$\frac{1}{3-x} =\frac13 \left(1 + \frac x3 + \frac {x^2}{9} + \frac {x^3}{27} +\> ... \right)  $$
Then, together,
$$\frac{1}{x-1} +\frac{1}{x-3} = -(1+\frac13) - (1+\frac19)x - (1+\frac{1}{27})x^2
-  (1+\frac{1}{81})x^3 -\> …$$
or, in the compact form,
$$\frac{1}{x-1} +\frac{1}{x-3} =- \sum\limits_{n=0}^\infty 
\left(1+\frac{1}{3^{n+1}}\right)x^n$$
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