Can we get a series for any rational function?

Can we get a series for this centered at \(1\)?

If we can get a series centered at \(1\), then the interval of convergence could only go out to \(2\). That’s a radius of \(1\). The interval of convergence would be \((0,2)\).

Our first hurdle is that we know how to get series for expressions that look like \(\frac {1}{1-something}\). Therefore, we need to split the formula for \(R(x)\) to look like

If we separate out the parts like \(\frac {A}{x+3}\) and \(\frac {B}{x-2}\), then no singularities should be left. That suggests that the remaining stuff is a polynomial.

We can work backwards. We’ll start by just combining the two fraction templates and see how close we get.

This will work provide

• \(A+B = -1\)
• \(3B-2A = 5\)

The first equation tells us that \(A = -1 - B\). Substituting that into the second equation gives us

\begin{align*} 3B-2A & = 5 \\ 3B-2(-1 -B) & = 5 \\ 2 + 5B & = 5 \\ 5B & = 3 \\ B & = \frac {3}{5} \end{align*}

\(A = -1 - \frac {3}{5} = \frac {-8}{5}\)

Now, we need to rewrite the fractions in the form \(\frac {1}{1 - something}\) where \(something(1) = 0\).

We can substitute these into the formula.

And, our series

Not a Geometric series, but a sum of two geometric series.