Power series of $\frac{1}{(x-1)(x-2)}$ at $x=0$

Question

I have to find power series of $\frac{1}{(x-1)(x-2)}$ at $x=0$ and give its radius of convergence.

I have

$$\frac{1}{(x-1)(x-2)}=\frac{1}{x-2}-\frac{1}{x-1}=\frac{1}{1-x}-\frac{1}{2-z}=\frac{1}{-2-i-x}-\frac{1}{-1-i-x}=\frac{1}{-2-i}\frac{1}{1-\frac{x}{-2-i}}-\frac{1}{-1-i}\frac{1}{1-\frac{x}{-1-i}}$$

But how do I continue with that?

Answer 1

You have that $$ \begin{gathered} f(x) = \frac{1} {{1 - x}} - \frac{1} {{2 - x}} \hfill \\ = \frac{1} {{1 - x}} - \frac{1} {2}\frac{1} {{1 - \frac{x} {2}}} = \hfill \\ = \sum\limits_{n = 0}^{ + \infty } {x^n } - \frac{1} {2}\sum\limits_{n = 0}^{ + \infty } {\frac{{x^n }} {{2^n }}} = \hfill \\ \hfill \\ = \sum\limits_{n = 0}^{ + \infty } {\left( {1 - \frac{1} {{2^{n + 1} }}} \right)x^n } \hfill \\ \end{gathered} $$ Since the first series has convergence radius $r_1=1$ while the second series has convergence radius $r_2$=2, your series has convergence radius $r=1$. The convergence radius is always the distance of the center of the series to the closest singularities. In our case the distance between $x_0=0$ and $x=1$.

Answer 2

You use the fact that\begin{align}\frac1{(x-1)(x-2)}&=\frac1{1-x}-\frac1{2-x}\\&=\frac1{1-x}-\frac12\cdot\frac1{1-\frac x2}\\&=(1+x+x^2+x^3+\cdots)-\frac12\left(1+\frac x2+\frac{x^2}{2^2}+\frac{x^3}{2^3}+\cdots\right)\\&=\left(1-\frac12\right)+\left(1-\frac1{2^2}\right)x+\left(1-\frac1{2^3}\right)x^2+\left(1-\frac1{2^4}\right)x^3+\cdots\end{align}