How to express $f(x)=\frac{1}{x-1} + \frac{1}{x-3}$ as a power series?

Question

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.

Answer 1

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,

$$\frac{1}{x-1} +\frac{1}{x-3} =- \sum\limits_{n=0}^\infty \left(1+\frac{1}{3^{n+1}}\right)x^n$$

Answer 2

As written in the comments for your question, for $\vert r\vert<1$ we know that:

$$ \sum\limits_{n=0}^\infty r^n= \frac{1}{1-r} $$

You can rewrite for all $a\neq 0$:

$$ \frac{1}{x-a}=-\frac{1}{a} \cdot \frac{1}{1-\frac{x}{a}}= -\frac{1}{a} \sum\limits_{n=0}^\infty \Big( \frac{x}{a} \Big)^n= -\sum\limits_{n=0}^\infty \frac{x^n}{a^{n+1}} $$

Which is true for all $\vert x\vert<\vert a\vert$. And you can apply that to your fractions.